Question

the sum of the third term and the seventh term of an ap is 6 and their product is 8 find the sum of the first 16 terms of the AP ​

Answer 1

remaining answer can be done by ypu

Answer 2

$\underline{ \large \sf \longmapsto Question:-}$

•The sum of the third term and the seventh term of an ap is 6 and their product is 8 find the sum of the first 16 terms of the AP .

$\underline{\large \longmapsto \sf Concept:-}$

Here, the concept of sum of nth term of A•P has been used to find out the required solution.

$\underline { \large\longmapsto \sf Solution:- }$

•Let the first term and the common difference of A.P be a and d respectively.

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•According to question

3rd term +7th term =6

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•Using below formula to find the nth term we get,

$\longrightarrow \sf a _{n} = a + (n - 1)d$

Now,

$\: \: \: \: \implies \sf \{a + (3 - 1)d \} + \{a + (7 - 1)d \} = 6$

$\: \: \: \implies \sf(a + 2d) + (a + 6d) = 6$

$\: \: \: \sf \implies 2a + 8d = 6$

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•Dividing throughout 2 on both side ,we get;

$\: \: \: \sf \implies \:a + 4d = 3 \: \: ....(1)$

and 3rd term×7th term=8

$\: \: \: \sf \implies(a + 2d)(a + 6d) = 8$

$\: \: \: \implies \sf(a + 4d - 2d)(a + 4d + 2d ) = 8$

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•Using equation (1) we get;

$\: \: \: \sf \implies(3 - 2d)(3 + 2d) = 8$

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•Using identity =a²-b²=(a+b)(a-b),we get;

$\: \: \: \implies \sf (3)^{2} - (2d)^{2} = 8$

$\: \: \: \implies \sf9 - 4d^{2} = 8$

$\: \: \: \implies \sf 4d^{2} = 9 - 8$

$\: \: \: \implies \sf 4d^{2} = 1$

$\: \: \: \implies \sf {d}^{2} = \dfrac{1}{4}$

$\: \: \: \implies \sf \: d = \sqrt{ \dfrac{1}{4} }$

$\: \: \: \implies \sf d = \pm \dfrac{1}{4}$

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$\bf \underline{Case \longrightarrow 1 } : when \: d = \dfrac{1}{2}$

Then from (1)

$\: \: \: \sf a + 4 \bigg( \dfrac{1}{2} \bigg) = 3$

$\: \: \: \implies \sf a + 2 = 3$

$\: \: \: \implies \sf \: a = 3 - 2 = 1$

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Therefore, Sum of first 16th term of A•P

$\implies \sf S_{16}$

$\implies \sf \dfrac{16}{2} \{2a + (16 - 1)d \}$

$\implies \sf 8(2a + 15d)$

$\implies \sf 8 \bigg \{2(1) + 15 \bigg( \dfrac{1}{2} \bigg) \bigg \}$

$\implies \sf \: 8 \bigg(2 + \dfrac{15}{2} \bigg)$

$\implies \sf \: 8 \bigg( \dfrac{9}{2} \bigg)$

$\implies \sf76$

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$\bf \underline{Case \longrightarrow2} : when \: d = - \dfrac{1}{2}$

Then from (1),

$\sf \: \: \: a + 4 \bigg( - \dfrac{1}{2} \bigg) = 3$

$\implies \sf a - 2 = 3$

$\implies \sf a = 3 + 2 = 5$

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Therfore,Sum of first sixteen of A•P

$\implies \sf S_{16}$

$\implies \sf \dfrac{16}{2} \{2a + (16 - 1)d \}$

$\implies \sf 8(2a + 15d)$

$\implies \sf 8 \bigg \{2(5) + 15 \bigg( - \dfrac{1}{2} \bigg) \bigg \}$

$\implies \sf8 \bigg(10 - \dfrac{15}{2} \bigg)$

$\implies \sf8 \bigg( \dfrac{5}{2} \bigg)$

$\implies \sf20$

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Therefore, the sum of first sixteen terms of A•P are 76 or 20.

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