JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Sequences and Series

Suppose

a_{1}, a_{2}, \dots, a_{n}, \dots

be an arithmetic progression of natural numbers. If the ratio of the sum of the first five terms of 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 -

Answer: (B)

Solution

Given

\frac{S _{5}}{S _{9}} = \frac{5}{17}

\frac{\frac{5}{2} ( 2 a + 4 d )}{\frac{9}{2} ( 2 a + 8 d )} = \frac{5}{17}

\frac{5 ( a + 2 d )}{9 ( a + 4 d )} = \frac{5}{17}

17 (a + 2 d) = 9 (a + 4 d)

17 a + 34 d = 9 a + 36 d

8 a = 2 d

d = 4 a

Now

a_{15} = a + 14 d = a + 56 a = 57 a

.

Given

110 < 57 a < 120

, so

a = 2

(the only natural number satisfying this).

Therefore

d = 8

S_{10} = \frac{10}{2} (2 a + 9 d) = 5 (4 + 72) = 5 \times 76 = 380

. Answer: (B)

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option