Question

The 14th term of an AP is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 terms.

Answer 1

the term14 of an AP is twice term=8

if the term is =6

then find the-8

sum of the first term =20

8×6=

Answer 2

Solution:-

•Let a and d be the first term and common difference of an AP a,a+d,a+2d...........respectively .

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It is given that

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$\sf \boxed{ \sf \:a_{14} = 2a _{8}}$

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We know that , $\sf a_n=a+(n-1)d$

$: \implies \sf \: \: a(14 - 1)d = \: 2(a + (8 - 1)d)$

$:\implies \sf\: \: a + 13d = \: 2(a + 7d)$

$: \implies \sf \: \: a + 13d = \: 2a + 14d$

$: \implies \sf \: \: a - 2a = \: 14d - 13d$

$: \implies \bf \: \: a = - d \: \: \: ....(1)$

Also,it is given that

$\boxed{ \sf a _{6} = - 8}$

$: \implies \sf \: \: a + (6 - 1)d = - 8$

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Using equation (1) we get,

$: \implies \sf \: \: - d + 5d= - 8$

$: \implies \sf \: \: 4d = - 8$

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

Substituting value of d=-2 in (1) ,

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

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Therefore, The A.P is

2,2+(-2),2+2(-2),2+3(-2).....

or 2,0,-2,-4........

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Now, we know that Sum of n terms of AP,

$\sf S_n= \dfrac{n}{2} [2a+(n-1)d]$

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Sum of first 20 terms of AP

$: \implies \sf S_{20}= \dfrac{20}{2} [2×2+(20-1)(-2)]$

$: \implies \sf S_{20}=10[4+(-38)]=10×(-34) \\ \sf= -340 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

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Hence, the sum of first 20 terms is -340