Question

the sum of 3 and 7th term of AP is 6 and the product is 8. find the sum of first sisteen terms of the AP.

I want ansnow it.self pls guys.
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Answer 1

The sum of first 16 term is = $\boxed{\sf \: 76}$

GIVEN :

• The sum of 3rd and 7th term of AP is 6.
• Product of the terms is 8

TO FIND : Sum of first 16 terms of the AP.

SOLUTION :

$\boxed{\sf\: A_n = a +(n-1) \times d }$

$\longrightarrow \sf\: Third\:term = a+2d \\ \\ \longrightarrow \sf\: Seventh\: term = a +6d$

$\bold{\underline{Condition\:1}}$

$\sf\: \dag Sum\:is\:6$

$\longrightarrow \sf\: a+2d+a+6d=6\\ \\ \longrightarrow \sf\: 2a+8d=6\\ \\ \longrightarrow \sf\:a+4d=3\\ \\ \longrightarrow \sf\:a=3-4d\:\:\:\:\:(1)$

$\bold{\underline{Condition\:2}}$

$\sf\: \dag Product\:is\:8$

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

From (1), putting the value of a in (2)

$\longrightarrow \sf\: (3-4d+2d)\times(3-4d+6d)=8\\ \\ \longrightarrow \sf\:(3-2d)\times(3+2d) = 8\\ \\ \longrightarrow \sf\:(3+2d)\times(3-2d)=8\\ \\ \longrightarrow \sf\:Using Identity (a+b)\times(a-b)={a}^{2}-{b}^{2}\\ \\ \longrightarrow \sf\:{3}^{2}-{(2d)}^{2}=8\\ \\ \longrightarrow \sf\:9-4{d}^{2}=8\\ \\ \longrightarrow \sf\:4{d}^{2}=1\\ \\ \longrightarrow \sf\:{d}^{2}=\frac{1}{4}\\ \\ \longrightarrow \sf\:d=\sqrt{\frac{1}{4}}=\frac{1}{2}\\ \\ \sf\:from\:(1)\\ \\ \longrightarrow \sf\:a=3-4d\\ \\ \longrightarrow \sf\:a=3-2=1$

$\sf\: \dag Sixteenth\:term$

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

$\longrightarrow \sf\: S_{16}=\frac{16}{2}(2+(16-1)\frac{1}{2}$$\longrightarrow \sf\:S_{16}=\frac{16}{2}(2+\frac{15}{2})\\ \\ \longrightarrow \sf\: S_{16}= 8 \times \frac{19}{2}\\ \\ \longrightarrow \sf\:S_{16}=76$

$\boxed{\sf\:Sixteenth\:term\:=\:76}$