Question

How many terms are there in an A.P. whose first term and 6th term are -12 and 8 respectively and sum of all it's terms is 120

Answer 1

Answer:

$S_{120}=27120$

Step-by-step explanation:

$Let \: a \: and \: d \: are \\first\:term \:and \: common\: difference\:of \: an \:A.P$

$We \: know \: that ,\\\boxed {n^{th}\: term =a_{n}=a+(n-1)d}$

According to the problem given,

$First\:term =a= -12$

$6^{th}\:term = 8$

$\implies a+5d = 8$

$\implies -12+5d = 8$

$\implies 5d = 8+12$

$\implies 5d = 20$

$\implies d = \frac{20}{5}$

$\implies d = 4$

$Now, \\Sum \:of \: n \:terms (S_{n})=\frac{n}{2}[2a+(n-1)d]$

$Here, a = -12 , \:d=4,\:n=120$

$S_{120}=\frac{120}{2}[2\times (-12)+(120-1)4]$

$=60(-24+119\times 4)$

$=60(-24+476)$

$=60\times 452$

$=27120$

Therefore,

$S_{120}=27120$

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Answer 2

Answer:

sn=27120

Step-by-step explanation:

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