23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by $1.$ Find the ${{15}^{th}}$ term. - Noon Academy

23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by $1.$ Find the ${{15}^{th}}$ term.

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

23. The eighth term of an A.P is half of its second term and the eleventh term exceeds one third of its fourth term by $1.$ Find the ${{15}^{th}}$ term.

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

Solution:

Given, an A.P in which,

${{a}_{8}}=1/2\left( {{a}_{2}} \right)$

${{a}_{11}}=1/3\left( {{a}_{4}} \right)+1$

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

⇒ ${{a}_{8}}=1/2\left( {{a}_{2}} \right)$

$a+7d=1/2\left( a+d \right)$

$2a+14d=a+d$

$a+13d=0$ …… $\left( i \right)$

And, ${{a}_{11}}=1/3\left( a+3d \right)+1$

$a+10d=1/3\left( a+3d \right)+1$

$3a+30d=a+3d+3$

$2a+7d=3$ …… $\left( ii \right)$

Solving $\left( i \right)$ and $\left( ii \right),$ by $\left( ii \right)-2x\left( i \right)$ ⇒

$2a+27d-2\left( a+13d \right)=3-0$

$d=3$

Putting $d$ in $\left( i \right)$ we get,

$a+13\left( 3 \right)=0$

$a=-39$

Thus, the ${{15}^{th}}$ term ${{a}_{15}}=-39+14\left( 3 \right)=-39+42=3$

i

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