11. The ${{26}^{th}}$, ${{11}^{th}}$ and the last term of an A.P. are $0,3$ and $-1/5,$ respectively. Find the common difference and the number of terms. - Noon Academy

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

Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma

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

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Given,

${{a}_{26}}=0,$ ${{a}_{11}}=3$ , a₁₁= 3 and ${{a}_{n}}$ (last term) $=-1/5$ of an A.P.

We know that, ${{n}^{th}}$ term ${{a}_{n}}=a+\left( n-1 \right)d$

Then,

${{a}_{26}}=a+\left( 26-1 \right)d$

⇒ $a+25d=0$ ….. $\left( 1 \right)$

And,

$a{}_{11}=a+\left( 11-1 \right)d$

⇒ $a+10d=3$ …… $\left( 2 \right)$

Solving $\left( 1 \right)$ and $\left( 2 \right)$ ,

$\left( 1 \right)$ – $\left( 2 \right)$ ⇒

$a+25d-\left( a+10d \right)=0-3$

$15d=-3$

⇒ $d=-1/5$

Using $d$ in $\left( 1 \right)$ , we get

$a+25\left( -1/5 \right)=0$

$a=5$

Now, given that the last term ${{a}_{n}}=-1/5$

⇒ $5+\left( n-1 \right)\left( -1/5 \right)=-1/5$

$5+-n/5+1/5=-1/5$

$25-n+1=-1$

$n=27$

Therefore, the A.P has $27$ terms and its common difference is $-1/5.$

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