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In an AP Given a3 = 15, S10 = 125, find d and a10. - CBSE Class 10 - Mathematics

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Question

In an AP

Given a3 = 15, S10 = 125, find d and a10.

Solution

Given that, a3 = 15, S10 = 125

As an = a + (n − 1)d,

a3 = a + (3 − 1)d

15 = a + 2d ... (i)

Sn = `n/2 [2a + (n - 1)d]`

S10 = `10/2 [2a + (10 - 1)d]`

125 = 5(2a + 9d)

25 = 2a + 9d ... (ii)

On multiplying equation (i) by (ii), we get

30 = 2a + 4d ... (iii)

On subtracting equation (iii) from (ii), we get

−5 = 5d

d = −1

From equation (i),

15 = a + 2(−1)

15 = a − 2

a = 17

a10 = a + (10 − 1)d

a10 = 17 + (9) (−1)

a10 = 17 − 9 = 8

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Textbook for Class 10 (2019 to Current)
Chapter 5: Arithmetic Progressions
Ex. 5.30 | Q: 3.04 | Page no. 112
Solution In an AP Given a3 = 15, S10 = 125, find d and a10. Concept: Sum of First n Terms of an AP.
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