Question

the sum of first n terms of ap is 155 and the sum of first 9 terms is 126, find 10th term​

Answer 1

Given:

The sum of the first 10 terms of ap is 155 and the sum of the first 9 terms is 126, find the 10th term​?

To find:

The 10th term

Solution:

We know,

$\boxed{\bold{S_n = \frac{n}{2}[2a + (n - 1)d] }}$

where n = no. of terms, a = first term, d = common difference and Sₙ = sum of n terms

Based on the formula above, we get

$S_1_0 = \frac{10}{2} [2a + (10 - 1)d] = 155$

$\implies 5 [2a + 9d] = 155$

$\implies 2a + 9d = 31$ . . . (1)

and

$S_9 = \frac{9}{2} [2a + (9 - 1)d] = 126$

$\implies 2a + 8d = 28$ . . . (2)

On subtracting equation (2) from (1), we get

2a + 9d = 31

2a + 8d = 28

-    -         -

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  d = 3

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On substituting d = 3 in equation (1), we get

$2a + (9\times 3) = 31$

$\implies 2a + 27 = 31$

$\implies 2a = 4$

$\implies a = 2$

We know,

$\boxed{\bold{a_n = a + (n-1)d}}$

where aₙ = last term of an A.P.

Therefore,

The 10th term of the A.P. is,

= a₁₀

= 2 + (10 - 1)3

= 2 + (9 × 3)

= 2 + 27

= 29

Thus, the 10th term of the A.P. is → 29.

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Also View:

The ratio of the 6th and 8th terms of an A.P is 7:9. find the ratio of 9th term to 13th term?

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Answer 2

I need step by step expla ation