Question

which term of the A.P. 11,8,5,...is -136?

Answer 1

SOLUTION :

Given : first term (a) = 11 , common difference (d) = 8 - 11 = -3

Let -140 be the nth term

an = a + (n - 1) d

-140 = a+(n-1)d

-140 = 11+(n-1) (-3)

-140 = 11 -3n +3

-140 = 14 - 3n

3n = 140 +14

3n=154

3n= 154

n = 154/3

n = 51.333…..

But n should be an integer

Hence -140 is not the term of the given AP

Answer 2

An= -136

a = 11

d = -3

$an = a + (n - 1)d \\ - 136 = 11 + (n - 1)( - 3) \\ - 136 - 11 = (n - 1)( - 3) \\ - 147 = (n - 1)( - 3) \\ \frac{ - 147}{- 3} = (n - 1) \\ 49 + 1 = n \\ 50 = n$

#hope it help uh.

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