Question

The sum of first 9 term of an a.p is 162.The Ratio of Its 6th term to its 13th term is 1:2.Find the first and 15th terms of the A.P

Answer 1

Step-by-step explanation:

hey dear...

Answer...

❤✌please refer to the attachment❤✌

✌hope it helps you✌

Answer 2

Given:-

• The sum of first 9 term of an A.P. is 162. ($\bold{\sf{S_9\:=\:162}})$
• The Ratio of Its 6th term to its 13th term is 1:2. ($\bold{\sf{a_{16}\::\:a_{13}\:=\:1\::\:2}}$)

Find:-

First and fifth term of an A.P. ($\bold{\sf{a\:and\:a_{15}}}$)

Solution:-

$\sf{S_n\:=\:\frac{n}{2}[2a\:+\:(n\:-\:1)d}$

Substitute the known values in above formula

=> $\sf{162\:=\:\frac{9}{2}[2a\:+\:(9\:-\:1)d]}$

=> $\sf{162\:=\:\frac{9}{2}[2a\:+\:8d]}$

=> $\sf{162\:=\:\frac{9}{2}\:\times\:2[a\:+\:4d]}$

=> $\sf{162\:=\:9[a\:+\:4d]}$

=> $\sf{\frac{162}{9}\:=\:a\:+\:4d}$

=> $\sf{18\:=\:a\:+\:4d}$

=> $\sf{a\:=\:18\:-\:4d}$ __ (eq 1)

$\sf{a_{6}\::\:a_{13}\:=\:1\::\:2}$

=> $\sf{ \frac{ a_{6} }{a_{13}} \:=\:\frac{1}{2}}$

=> $\sf{ \frac{ a\:+\:(6\:-\:1)d}{a\:+\:(13\:-\:1)d} \:=\:\frac{1}{2}}$

=> $\sf{2(a\:+\:5d\:=\:a\:+\:12d}$

=> $\sf{2a\:+\:10d\:=\:a\:+\:12d}$

=> $\sf{a\:=\:2d}$

=> $\sf{18\:-4d\:=\:2d}$

=> $\sf{6d\:=\:18}$

=> $\sf{d\:=\:3}$

Substitute value of d in (eq 1)

=> $\sf{a\:=\:18\:-\:12}$

=> $\sf{a\:=\:6}$

$\therefore\:\sf{a_{15}\:=\:a\:+\:(15\:-\:1)d}$

=> $\sf{6\:+\:14(3)}$

=> $\sf{6\:+\:42}$

=> $\sf{48}$

•°• First term is 6 and fifth term is 48.