Question

In an A.P if S10 = 35 and S9 = 28 find a10.​

Answer 1

Given:

S₁₀ = 35

S₉ = 28

To find:

a₁₀

Solution:

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

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

35 = 5(2a + 9d)

7 = 2a + 9d - (I)

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

28 = (9/2)(2a + 8d)

28 = 9(a + 4d)

28 = 9a + 36d - (II)

On solving equation (I) and (II)

a = 0

d = 7/9

aₙ = a + (n - 1)d

a₁₀ = 0 + (10 - 1)(7/9)

a₁₀ = 9(7/9)

a₁₀ = 7

So, a₁₀ is equal to 7.

Answer 2

SOLUTION

GIVEN

$\sf{In \: an \: AP \: S_{10 }= 35 \: and \: S_9 = 28}$

TO DETERMINE

$\sf{a_{10 }}$

EVALUATION

We know that if in an arithmetic progression

$\sf{Sum \: of \: n \: terms = S_n}$

$\sf{n \: th \: term = a_n}$

Then

$\sf{a_n = S_n - S_{n-1}}$

Putting n = 10 we get

$\sf{a_{10} = S_{10} - S_9}$

$\sf{ \implies \: a_{10} = 35 - 28}$

$\sf{ \implies \: a_{10} = 7}$

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