Question

Suppose a 1, a 2, … a n, … be an arithmetic progression of natural numbers. If the ration of the sum of first five terms to the sum of first nine terms of the progression is 5 : 17 and 110 < a 15 < 120, then the sum of the first ten terms of the progression is equal to

• A. 290
• B. 380
• C. 460
• D. 510

Answer 1

Answer: (B)

380

Answer 2

The correct answer is (B):
∵ a 1, a 2, … a n … be an A.P of natural numbers and
$\frac{S_5}{S_9} =\frac{ 5}{17}$
⇒ $\frac{\frac{5}{2}[2α_1+4d]}{\frac{9}{2}[2α_1+8d]} = \frac{5}{17}$
⇒ 34α1+68d
= 18α1+72d
∴ d = 4α1
And 110<α15<120
∴ 110<α1+14d<120
⇒ 110<57α1<120
∴ α1 = 2(∵ αi ∈N)
d = 8
∴ S10 = 5[4+9×8]
= 380