The sum of $4^{t h}$ and $8^{t h}$ form of an $A P$ is $24$ and the sum of its $6^{t h}$ and $10^{t h}$ terms is $44$ . Find the sum of first $10^{t h} t e r m s$ of $A P$ .

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Solution

$T_{n} = a + (n - 1) d$

$T_{4} = a + (4 - 1) d = a + 3 d$

Similarly

$T_{8} = a + 7 d, T_{6} = a + 5 d, T_{10} = a + 9 d$

Given,

$a + 3 d + a + 7 d = 24$

$\Rightarrow 2 a + 10 d = 24$

$\Rightarrow a + 5 d = 12$ ...(1)

$a + 5 d + a + 9 d = 44$

$\Rightarrow 2 a + 14 d = 44$

$\Rightarrow a + 7 d = 22$ ...(2)

(2) - (1) gives

$2 d = 10 \Rightarrow d = 5$

From (2)

$a = 22 - 7 (5) = - 13$

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

$S_{10} = \frac{10}{2} [2 (- 13) + (10 - 1) 5]$

$= 5 [- 26 + 45]$

$= 95$

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The sum of the $4^{th}$ and $8^{th}$ term of an A.P is $24$ and the sum of its $6^{th}$ and $10^{th}$ terms is $44$ . Find the first term?

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