Question

in an ap if S5 +S7 = 167 and S10=235 find the AP where SN denotes the sum of its first n terms

Answer 1

now u may solve these 2 equation by elemination method

Answer 2

Answer: 1,6,11,16......

Step-by-step explanation:

S10=10/2 (2a+9d)=235

5(2a+9d)=235

10a+45d=235

Dividing whole by 5

2a+9d=47

Multiplying by 6

12a+54d=282 ~ {1}

S5+S7=5/2 [2a+(5-1)d]+ 7/2 [2a+(7-1)d]

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

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

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

12a+31d=167~{2}

Subtracting {2} from {1}, we get,

23d=115

D=115/23

D=5

2a +9d = 47

2a+ 9*5=47

2a=49-47

2a=2

A=1

A.P., is 1,6,11,16…