Question

In an a.p. if s5+ s7=16 and s10=20th end find the a.p.

Answer 1

Solution:

Given: In the A.P $S_{5} +S_{7}= 16$ 
$(a_1+a_2+a_3+a_4+a_5)+(a_1+a_2+a_3+a_4+a_5+a_6+a_7)=16$
as,we know, $a_n=a+(n-1)d$
$[a+(a+d)+(a+2d)....(a+4d)]+[a+(a+d)+(a+2d)....(a+6d)]=16$
$[5a+10d]+[7a+21d]=16$
$12a+31d=16.....(i)$

$S_{10}=20$ which means,
$[a_1+a_2+a_3...a_{10}] =20$
[tex][a+(a+d)+(a+2d)...(a+9d)]=20 [/tex]
$10a+45d=20$            (when divided by 5)
$2a+9d=4...(ii)$

(i) - 6x(ii) = [tex]12a+31d-6(2a+9d)=16-6(4) [/tex]
                     $12a+31d-12a-54d=16-24$
                                            $-23d=-8$
                                             [tex]d=8/23 ...(iii)[/tex]
 substituting value of d in (ii)
                  $2a+9d=4$
           $2a+9(8/23)=4$
           $2a+(72/23)=4$
                          [tex]2a=4-(72/23) [/tex]
                          $2a=92-72/23$
                          $2a=20/23$
                          ∴$a= 10/23$
                                       Hope it Helps

Answer 2

Answer:

S5+S7=5/2(2a+4d) +7/2(2a+6d)=167

=5a+10d+7a+21d=167

=12a+31d=167   [let this be eqn 1]

similarly,

from S10=235, we get 2a+9d=47 {let this be equation 2}

by solving eqn 1 and 2, we get a=1 and d=5

therefore the ap is 1,6,11........

Step-by-step explanation: