Question

The first and last term of an Arithmetic progression is -12 and 40 if the sum of the sequence is 196 find the number of terms

Answer 1

a (first term) = - 12

l (last term) = 40

Sn (Sum) = 196

n (NO. Of terms) =?

Sn = n/2(a + l)

196 = n/2(-12 + 40)

392 = 28n

n = 14

Hence total terms are 14.

Thanks

Answer 2

some important concepts :

T 1 = a = first term

t 2 = second term

d = common difference

n = no. of terms

t n = last term

s = sum

s n = sum of all terms in ap

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here

t 1 = -12

t n = 40

s n = 196

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let's use the formula :

$s \: n \: = \frac{n}{2} ( \: t \: 1 \: + \: t \: n \: ) \\ \\ 196 = \frac{n}{2} \times ( - 12 + 40 \: ) \\ \\ 196 \times 2 = n \times (28) \\ \\ n = \frac{196 \times 2}{28} \\ \\ n = 7 \times 2 \\ \\ n = 14$
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$\bold{no. \: of \: terms \: is \: 14 \: }$

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