Question

• In an AP, if S5+S7=167 and S10=235, then find S4 , where Sn denotes the sum of its first n terms

Answer 1

Step-by-step explanation:

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let a and d be the first term and the common differences of AP.

= sum of n terms =

Sn=n/2[2a+(n-1)d]

we have..

= S5+S7=167

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

=5(a+2d)+7(a+3d)=167

=12a+31d=167...(1)

again...

S10=235

= 10/2(2a+9d)=235

=2a+9d=47........(2)

solving equation1 and equation2.....

we get...

a=1

d=5

=AP=1,6,11,16......

S4=4/2(2a+3d)

= 2×(2+15)

=2×17

=34

Answer 2

Step-by-step explanation:

$\huge\boxed{\underline{\mathfrak{\red{ answer} } } }$

let a and d be the first term and the common differences of AP.

= sum of n terms =

Sn=n/2[2a+(n-1)d]

we have..

= S5+S7=167

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

=5(a+2d)+7(a+3d)=167

=12a+31d=167...(1)

again...

S10=235

= 10/2(2a+9d)=235

=2a+9d=47........(2)

solving equation1 and equation2.....

we get...

a=1

d=5

=AP=1,6,11,16......

S4=4/2(2a+3d)

= 2×(2+15)

=2×17

=34