Question

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The sum of 4th and 8th term of an AP is 24 and the sum of its 6th and 10th terms is 44. Find the sum of 10 terms of the AP.

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Answer 1

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Answer 2

Answer

The sum of 10 terms of the AP is 95

$\bf \large \underline{Given : }$

• The sum of 4th and 8th term of an AP is 24 and sum of its 6th and 10th terms is 44

$\bf \large \underline{To \: Find : }$

• The sum upto 10 terms of the AP

$\bf \large \underline{Formula \: to \: be \: used : }$

If the first term and common difference of an AP is a and d respectively then ,

$\bullet \: \: \: \tt a _{n} = a + (n - 1)d$

$\tt \bullet \: \: \: S_{n} = \dfrac{n}{2} \{ 2a + (n - 1)d \}$

$\bf \large \underline{Solution :}$

We are given ,

$\sf \implies 4th \: term + 8th \: term= 24 \\ \\ \implies \sf a + 3d + a + 7d = 24 \\ \\ \sf \implies 2a + 10d = 24 \\ \\ \implies \sf a + 5d = 12 \longrightarrow (1)$

Again ,

$\sf \implies 6th \: term+ 10th \: term= 44 \\ \\ \implies \sf a + 5d + a + 9d = 44 \\ \\ \sf \implies 2a + 14d = 44 \\ \\ \implies \sf a + 7d = 22 \longrightarrow(2)$

Subtracting (1) from (2) we have ,

$\sf \implies a + 7d - a - 5d = 22 - 12 \\\\ \sf\implies 2d = 10 \\\\ \sf\implies d = 5$

Putting the value of ‘d’ in (1) we have ,

$\sf\implies a + 5\times 5 = 12 \\\\ \sf\implies a = 12 - 25 \\\\ \sf\implies a = -13$

Now sum upto 10 terms of the AP is :

$\sf \implies S_{10} = \dfrac{10}{2}\{ 2\times (-13) + (10-1)5 \} \\\\ \sf\implies S_{10}=5\{-26 + 45 \} \\\\ \sf\implies S_{10} = 5\times 19 \\\\ \sf\implies S_{10}=95$