CBSE Class 10 Mathematics chapter 5 - ARITHMETIC PROGRESSIONS - Exercise 5.3

CBSE Class 10 Mathematics chapter 5 - ARITHMETIC PROGRESSIONS - Exercise 5.3

Q1.	Find the sum of the following APs:
(i)	2, 7, 12, . . ., to 10 terms.
Solution.

(ii)	–37, –33, –29, . . ., to 12 terms.
Solution.

(iii)	0.6, 1.7, 2.8, . . ., to 100 terms.
Solution.

(iv)
Solution.

Q2.	Find the sums given below :
(i)
Solution.	Given – a = 7 l = an =  84 d = 10.5 — 7 = 3.5  (because 10½ = 10.5) n = ? S = ? We know that n = 1 + (an—a)/d n = 1 + (84—7)/3.5 n = 1 + 77 /3.5 n = 1+ 22 n = 23

(ii)	34 + 32 + 30 + . . . + 10
Solution.	Given – a = 34 l = an =  10 d = 32 — 34 = —2 n = ? S = ? We know that n = 1 + (an—a)/d n = 1 + (10—34)/(—2) n = 1 + (—24) /(—2) n = 1+ 12 n = 13

(iii)	–5 + (–8) + (–11) + . . . + (–230)
Solution.	Given – a = —5 l = an =  —230 d = (—8) — (—5) = —3 n = ? S = ? We know that n = 1 + (an—a)/d n = 1 + ((—230)—(—5))/(—3) n = 1 + (—230+5) /(—3) n = 1+ (—225)/(—3) n = 1 + 75 n = 76

Q3.	In an AP:
(i)	given a = 5, d = 3, an= 50, find n and Sn
Solution.	We know that n = 1 + (an—a)/d n = 1 + (50—5)/3 n = 1 + 45 /3 n = 1 + 15 n = 16

(ii)	given a = 7, a13 = 35, find d and S13
Solution.	We know that – an = a + (n—1)d For a13 a13 = a + (13—1)d a13 = a + 12d a13 — a = 12d

(iii)	given a12 = 37, d = 3, find a and S12
Solution.	We know that – an = a + (n—1)d For a12 a13 = a + (12—1)d a13 = a + 11d a  = a13 — 11d a = 37—11×3 a = 37—33 a = 4

(iv)	given a3 = 15, S10 = 125, find d and a10
Solution.	We know that – an = a + (n—1)d For a3 a3 = a + (3—1)d a3 = a + 2d a = a3 — 2d putting value when an = a3 = 15 a = 15—2d ……………… (i) We know that

(v)	given d = 5, S9 = 75, find a and a9
Solution.

(vi)	given a = 2, d = 8, Sn = 90, find n and an
Solution.

(vii)	given a = 8, an = 62, Sn = 210, find n and d
Solution.

(viii)	given an = 4, d = 2, Sn = –14, find n and a
Solution.

(ix)	given a = 3, n = 8, S = 192, find d
Solution.

(x)	given l = 28, S = 144, and there are total 9 terms. Find a
Solution.