Question

Find the 12th term from end of the ap,1,4,7, 88

Answer 1

$\sf\blue{Question}$

$\sf{Find \ the \ 12^{th} \ term \ from \ end \ of \ the \ AP}$

$\sf{1, \ 4, \ 7,...,88}$

_______________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ 12^{th} \ term \ from \ the \ end \ is \ 55.}$

$\sf\orange{Given:}$

$\sf{The \ given \ AP \ is}$

$\sf{\implies{1, \ 4, \ 7,...,88}}$

$\sf\pink{To \ find:}$

$\sf{12^{th} \ term \ from \ the \ last.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ AP \ is}$

$\sf{\implies{1, \ 4, \ 7,...,88}}$

$\sf{Here \ a=1, d=4-1=3, \ t_{n}=88}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{88=1+(n-1)3}}$

$\sf{\therefore{3(n-1)=88-1}}$

$\sf{\therefore{3(n-1)=87}}$

$\sf{\therefore{n-1=\frac{87}{3}}}$

$\sf{\therefore{n=29+1}}$

$\sf{\therefore{n=30}}$

$\sf{Second \ term \ from \ last=(n-1)^{th} \ term}$

$\sf{\therefore{12^{th} \ term\ from \ the \ last=(n-11)^{th} \ term}}$

$\sf{=(30-11)^{th} \ term=19^{th} \ term}$

$\sf{\therefore{t_{19}=1+(19-1)3}}$

$\sf{\therefore{t_{19}=1+18(3)}}$

$\sf{\therefore{t_{19}=1+54}}$

$\sf{\therefore{t_{19}=55}}$

$\sf\purple{\tt{\therefore{The \ 12^{th} \ term \ from \ the \ end \ is \ 55.}}}$

Answer 2

Answer:

The sequence is an A.P as common difference is same.

d = 3

tn = a + (n-1)×d

a = first term = 1

tn = last term = 88

common difference d = 3

substituting values

88 = 1 + (n-1)×3

88 = 1 + 3n-3

88 = -2+3n

90 = 3n

n = 90/3

n = 30.

There are 30 terms in A.P

t19 will be 12th term from end

t19 = 55

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