Title 02 Arithmetic Progressons (Part-1)(1).pdf ✅

ARITHMETIC PROGRESSION 2 CHAPTER ➢ DEFINITION ➢ When the terms of a sequence or series are arranged under a definite rule then they are said to be in a Progression. Progression can be classified into 5 parts as - (i) Arithmetic Progression (A.P.) (ii) Geometric Progression (G.P.) (iii) Arithmetic Geometric Progression (A.G.P.) (iv) Harmonic Progression (H.P.) (v) Miscellaneous Progression ➢ ARITHMETIC PROGRESSION (A.P.) ➢ Arithmetic Progression is defined as a series in which difference between any two consecutive terms is constant throughout the series. This constant difference is called common difference. If ‘a’ is the first term and ‘d’ is the common difference, then an AP can be written as a + (a + d) + (a + 2d) + (a + 3d) + ...... Note: If a,b,c, are in AP  2b = a + c General Term of an AP General term (nth term) of an AP is given by Tn = a + (n– 1) d Note : (i) General term is also denoted by  (last term) (ii) n (No. of terms) always belongs to set of natural numbers. (iii) Common difference can be zero, + ve or – ve. d = 0  then all terms of AP are same Eg. 2, 2, 2, 2, ......... d = 0 d = +ve  increasing AP Eg. 2 5 , 3, 2 7 , 4, 2 9 ,......... d = + 2 1 d = –ve  decreasing A.P. Eg. 57, 52, 47, 42, 37, ........ d = – 2 1 r th term from end of an A.P. If number of terms in an A.P. is n then Tr from end = Tn – (r – 1)d = (n – r + 1)th from beginning or we can use last term of series as first term and use ‘d’ with opposite sign of given A.P. Eg. : Find 26th term from last of an AP 7, 15, 23......., 767 consits 96 terms. Sol. Method : I r th term from end is given by = Tn – (r – 1) d or = (n – r + 1) th term from beginning where n is total no. of terms. m = 96, n = 26  T26 from last = T(96 – 26 + 1) from beginning = T71 from beginning = a + 70d = 7 + 70 (8) = 7 + 560 = 567 Method : II d = 15 – 7 = 8  from last, a = 767 and d = –8  T26 = a + 25d = 767 + 25 (–8) = 767 – 200 = 567. Sum of n terms of an A.P. The sum of first n terms of an A.P. is given by Sn = 2 n [2a + (n – 1) d] or Sn = 2 n [a + Tn ] Note : (i) If sum of n terms Sn is given then general term Tn = Sn – Sn–1 where Sn–1 is sum of (n – 1) terms of A.P.
(ii) n th term of an AP is linear in ‘n’ Eg. : an = 2 – n, an = 5n + 2........ Also we can find common difference ‘d’ from an or Tn : d = coefficient of n For an = 2 – n  d = –1 Ans. Verification : by putting n = 1, 2, 3, 4,......... we get AP : 1, 0, –1, –2,........  d = 0 – 1 = –1 Ans. & for an = 5n + 2 d = 5 Ans. (iii) Sum of n terms of an AP is always quadratic in ‘n’ Eg. : Sn = 2n2 + 3n. Eg. : Sn = 4 n (n + 1) we can find ‘d’ also from Sn. d = 2 (coefficient of n2 ) for eg. : 2n2 + 3n, d = 2(2) = 4 Verification Sn = 2n2 + 3n at n = 1 S1 = 2 + 3 = 5 = first term at n = 2 S2 = 2(2)2 + 3(2) = 8 + 6 = 14  second term = sum of first two terms.  second term = S2 – S1 = 14 – 5 = 9  d = a2 – a1 = 9 – 5 = 4 Eg. : Sn = 4 n (n + 1) Sn = 4 n 4 n 2 +  d = 2 2 1 4 1  =      Ans. ❖ EXAMPLES ❖ Ex.1 If the nth term of a progression be a linear expression in n, then prove that this progression is an AP. Sol. Let the nth term of a given progression be given by Tn = an + b, where a and b are constants. Then, Tn–1 = a(n – 1) + b = [(an + b) – a]  (Tn – Tn–1) = (an + b) – [(an + b) – a] = a, which is a constant. Hence, the given progression is an AP. Ex.2 Write the first three terms in each of the sequences defined by the following - (i) an = 3n + 2 (ii) an = n2 + 1 Sol.(i) We have, an = 3n + 2 Putting n = 1, 2 and 3, we get a1 = 3 × 1 + 2 = 3 + 2 = 5, a2 = 3 × 2 + 2 = 6 + 2 = 8, a3 = 3 × 3 + 2 = 9 + 2 = 11 Thus, the required first three terms of the sequence defined by an =3n +2 are 5, 8, and 11. (ii) We have, an = n2 + 1 Putting n = 1, 2, and 3 we get a1 = 12 + 1 = 1 + 1 = 2 a2 = 22 + 1 = 4 + 1 = 5 a3 = 32 + 1 = 9 + 1 = 10 Thus, the first three terms of the sequence defined by an = n2 + 1 are 2, 5 and 10. Ex.3 Write the first five terms of the sequence defined by an = (–1)n–1 . 2n Sol. an = (–1)n–1 × 2n Putting n = 1, 2, 3, 4, and 5 we get a1 = (–1)1–1 × 21 = (–1)0 × 2 = 2 a2 = (–1)2–1 × 22 = (–1)1 × 4 = – 4 a3 = (–1)3–1 × 23 = (–1)2 × 8 × 8 a4 = (–1)4–1 × 24 = (–1)3 × 16 = –16 a5 = (–1)5–1 × 25 = (–1)4 × 32 = 32 Thus the first five term of the sequence are 2, –4, 8, –16, 32. Ex.4 The nth term of a sequence is 3n – 2. Is the sequence an A.P. ? If so, find its 10th term. Sol. We have an = 3n – 2 Clearly an is a linear expression in n. So, the given sequence is an A.P. with common difference 3. Putting n = 10, we get a10 = 3 × 10 – 2 = 28 REMARK : It is evident from the above examples that a sequence is not an A.P. if its nth term is not a linear expression in n.
Ex.5 Find the 12th, 24th and nth term of the A.P. given by 9, 13, 17, 21, 25, ......... Sol. We have, a = First term = 9 and, d = Common difference = 4 [ 13 – 9 = 4, 17 – 13 = 4, 21 – 7 = 4 etc.] We know that the nth term of an A.P. with first term a and common difference d is given by an = a + (n – 1) d Therefore, a12 = a + (12 – 1) d = a + 11d = 9 + 11 × 4 = 53 a24 = a + (24 – 1) d = a + 23 d = 9 + 23 × 4 = 101 and, an = a + (n – 1) d = 9 + (n – 1) × 4 = 4n + 5 a12 = 53, a24 = 101 and an = 4n + 5 Ex.6 Which term of the sequence –1, 3, 7, 11, ..... , is 95 ? Sol. Clearly, the given sequence is an A.P. We have, a = first term = –1 and, d = Common difference = 4. Let 95 be the nth term of the given A.P. then, an = 95  a + (n – 1) d = 95  – 1 + (n – 1) × 4 = 95  – 1 + 4n – 4 = 95  4n – 5 = 95  4n = 100  n = 25 Thus, 95 is 25th term of the given sequence. Ex.7 Which term of the sequence 4, 9 , 14, 19, ...... is 124 ? Sol. Clearly, the given sequence is an A.P. with first term a = 4 and common difference d = 5. Let 124 be the nth term of the given sequence. Then, an = 124 a + (n – 1) d = 124  4 + (n – 1) × 5 = 124  n = 25 Hence, 25th term of the given sequence is 124. Ex.8 The 10th term of an A.P. is 52 and 16th term is 82. Find the 32nd term and the general term. Sol. Let a be the first term and d be the common difference of the given A.P. Let the A.P. be a1 , a2 , a3 , ..... an , ...... It is given that a10 = 52 and a16 = 82  a + (10 – 1) d = 52 and a + (16 – 1) d = 82  a + 9d = 52 ....(i) and,a + 15d = 82 ....(ii) Subtracting equation (ii) from equation (i), we get –6d = – 30  d = 5 Putting d = 5 in equation (i), we get a + 45 = 52  a = 7  a32 = a + (32 – 1) d = 7 + 31 × 5 = 162 and, an = a + (n – 1) d = 7 (n – 1) × 5 = 5n + 2. Hence a32 = 162 and an = 5n + 2. Ex.9 Determine the general term of an A.P. whose 7 th term is –1 and 16th term 17. Sol. Let a be the first term and d be the common difference of the given A.P. Let the A.P. be a1 , a2 , a3 , ....... an , ....... It is given that a7 = – 1 and a16 = 17 a + (7 – 1) d = – 1 and, a + (16 – 1) d = 17  a + 6d = – 1 ....(i) and,a + 15d = 17 ....(ii) Subtracting equation (i) from equation (ii), we get 9d = 18  d = 2 Putting d = 2 in equation (i), we get a + 12 = – 1  a = – 13 Now, General term = an = a + (n – 1) d = – 13 + (n – 1) × 2 = 2n – 15 Ex.10 If five times the fifth term of an A.P. is equal to 8 times its eight term, show that its 13th term is zero. Sol. Let a1 , a2 , a3 , ..... , an , .... be the A.P. with its first term = a and common difference = d. It is given that 5a5 = 8a8  5(a + 4d) = 8 (a + 7d)  5a + 20d = 8a + 56d  3a + 36d = 0  3(a + 12d) = 0  a + 12d = 0  a + (13 – 1) d = 0  a13 = 0 Ex.11 If the mth term of an A.P. be 1/n and nth term be 1/m, then show that its (mn)th term is 1.
Sol. Let a and d be the first term and common difference respectively of the given A.P. Then, n 1 = mth term  n 1 = a + (m – 1) d ....(i) m 1 = nth term  m 1 = a + (n – 1) d ....(ii) On subtracting equation (ii) from equation (i), we get n 1 – m 1 = (m – n) d  mn m − n = (m – n) d  d = mn 1 Putting d = mn 1 in equation (i), we get n 1 = a + mn (m −1)  a = mn 1  (mn)th term = a + (mn – 1) d = mn 1 + (mn – 1) mn 1 = 1 Ex.12 If m times mth term of an A.P. is equal to n times its nth term, show that the (m + n) term of the A.P. is zero. Sol. Let a be the first term and d be the common difference of the given A.P. Then, m times mth term = n times nth term  mam = nan  m{a + (m – 1) d} = n {a + (n – 1) d}  m{a + (m – 1) d} – n{a + (n – 1) d} = 0  a(m – n) + {m (m – 1) – n(n – 1)} d = 0  a(m – n) + (m – n) (m + n – 1) d = 0  (m – n) {a + (m + n – 1) d} = 0  a + (m + n – 1) d = 0  am+n = 0 Hence, the (m + n)th term of the given A.P. is zero. Ex.13 If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n). Sol Let a be the first term and d be the common difference of the given A.P. Then, p th term = q  a + (p – 1) d = q ....(i) q th term = p  a + (q – 1) d = p ....(ii) Subtracting equation (ii) from equation (i), we get (p – q) d = (q – p)  d = – 1 Putting d = – 1 in equation (i), we get a = (p + q – 1) n th term = a + (n – 1) d = (p + q – 1) + (n – 1) × (–1) = (p + q – n) Ex.14 If pth, qth and rth terms of an A.P. are a, b, c respectively, then show that (i) a (q – r) + b(r – p) + c(p – q) = 0 (ii) (a – b) r + (b – c) p + (c – a) q = 0 Sol. Let A be the first term and D be the common difference of the given A.P. Then, a = pth term  a = A + (p – 1) D ....(i) b = qth term  b = A + (q – 1) D ....(ii) c = rth term  c = A+ (r – 1) D ....(iii) (i) : We have, a(q – r) + b (r – p) + c (p – q) = {A + (p – 1) D} (q – r) + {A + (q – 1)} (r – p) + {A + (r – 1) D} (p – q) [Using equations (i), (ii) and (iii)] = A {(q – r) + (r – p) + (p – q)} + D {(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)} = A {(q – r) + (r – p) + (p – q)} + D{(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)} = A . 0 + D {p (q – r) + q (r – p) + r (p – q) – (q – r) – (r – p) – (p – q)} = A . 0 + D . 0 = 0 (ii) : On subtracting equation (ii) from equation (i), equation (iii) from equation (ii) and equation (i) from equation (iii), we get a – b = (p – q) D, (b – c) = (q – r) D and c – a = (r – p) D  (a – b) r + (b – c) p + (c – a) q = (p – q) Dr + (q – r) Dp + (r – p) Dq = D {(p – q) r + (q – r) p + (r – p) q} = D × 0 = 0 Ex.15 Determine the 10th term from the end of the A.P. 4, 9, 14, ........, 254. Sol. We have, l = Last term = 254 and, d = Common difference = 5, 10th term from the end = l – (10 – 1) d = l – 9d = 254 – 9 × 5 = 209.