10. Show that a1, a2 … , an , … form an AP where an is defined as below(i) an = 3+4n(ii) an = 9−5nAlso find the sum of the first 15 terms in each case.

(i) an = 3+4n a1 = 3+4(1) = 7 a2 = 3+4(2) = 3+8 = 11 a3 = 3+4(3) = 3+12 = 15 a4 = 3+4(4) = 3+16 = 19 We can see here, the common difference between the terms are; a2 − a1 = 11−7 = 4 a3 − a2 = 15−11 = 4 a4 − a3 = 19−15 = 4 Hence, ak + 1 − ak is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7. Now, we know, the sum of nth term is; Sn = n/2[2a+(n -1)d] S15 = 15/2[2(7)+(15-1)×4] = 15/2[(14)+56] = 15/2(70) = 15×35 = 525 (ii) an = 9−5n a1 = 9−5×1 = 9−5 = 4 a2 = 9−5×2 = 9−10 = −1 a3 = 9−5×3 = 9−15 = −6 a4 = 9−5×4 = 9−20 = −11 We can see here, the common difference between the terms are; a2 − a1 = −1−4 = −5 a3 − a2 = −6−(−1) = −5 a4 − a3 = −11−(−6) = −5 Hence, ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4. Now, we know, the sum of nth term is; Sn = n/2 [2a +(n-1)d] S15 = 15/2[2(4) +(15 -1)(-5)] = 15/2[8 +14(-5)] = 15/2(8-70) = 15/2(-62) = 15(-31) = -465