The $ 26^{\text {th }}, 11^{\text {th }} $ and the last term of an AP are 0,3 and $ -\frac{1}{5} $, respectively. Find the common difference and the number of terms.

AcademicMathematicsNCERTClass 10

Given:

The 26th, 11th and last term of an A.P. are $0, 3$ and $-\frac{1}{5}$, respectively.

To do:

We have to find the common difference and the number of terms.

Solution:

Let the first term, common difference and the number of terms of the given A.P. be $a, d$ and $n$ respectively.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{26}=a+(26-1)d$

$0=a+25d$

$a=-25d$.....(i)

$a_{11}=a+(11-1)d$

$3=a+10d$

$3=-25d+10d$     (From (i))

$3=-15d$

$d=\frac{3}{-15}$

$d=\frac{-1}{5}$....(ii)

Last term $l=a+(n-1)d$

$-\frac{1}{5}=-25(\frac{-1}{5})+(n-1)\frac{-1}{5})$     (From (i) and (ii))

$1=-25+(n-1)$

$1+25+1=n$

$n=27$

The common difference and the number of terms are $\frac{-1}{5}$ and $27$ respectively.

raja
Updated on 10-Oct-2022 13:27:34

Advertisements