Question

In an ap S5 + S7 is equal to 167 and S10 is 235 find the AP where SN denotes the first of n terms

Answer 1

S10=235
5(a+9d)=235
a+9d=47.......eq1
S5+S7=167
5/2(a+4d)+7/2(a+6d)=167
5a/2+10d+7a/2+21d=167
31d+6a=167.......eq2
On solving eq 1 and 2
a=2
d=5
Sn=n/2[a+(n-1)d]
Sn=n/2[2+(n-1)(5)]
Sn=n/2(2+5n-5)
Sn=n/2(5n-3)

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: