Question

can 152 be the term of the AP 12, 15, 18, 21,.......

Answer 1

here...
a=12
d=15-12=3

In terms of AP
An=a+(n-1)d
if n is a whole no: then 152 is a term of the given AP

so 152=12+(n-1)(3)
→152-12=(n-1)(3)
→140/3=n-1
→140/3 -1 =n
→n=137/3=45.666.. ≠ a whole no:

Hence 152 Cannot be a term on the AP.

nearest term would be 138+12=150 or 153.

Hope it helped...
Regards
Leukonov.

Answer 2

hii fellow!!
here,
a=12
d=15-12=3
$an = l = 152 \\ an = a + (n - 1)d \\ 152 = 12 + (n - 1)3 \\ 152 - 12 = (n - 1)3 \\ \frac{140}{3 } = n - 1 \\ 46.66 + 1 = n \\ 47.66 = n$
$47.66 = n$
but we know that n must not be a fraction or in decimals or negative
therefore,152 is not a term of this AP

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