Question

if s5+s7=167 or s10=235 then find Ap where sn denotes sum of its first n term

Answer 1

Answer:

Step-by-step explanation:

sn=n/2(2a+(n-1)d)                              s10=10/2(2a+9d)

s5+s7=167                                         s10=235

5/2(2a+4d)+7/2(2a+6d)=210             235/5=2a+9d

5(2a+4d)+7(2a+6d)=210*2                47=2a+9d.........(2)

10a+20d+14a+42d=420

24a+62d=420

12a+31d=210......(1)

from 1 and 2

12a+31d=210

(2a+9d=47)6

12a+54d=282

12a+31d=210 (by changing signs)

       23d=72

d=72/23          2a+9d=47                      

d=3.13             a=47- 28.17/2= 9.145

hence,

AP=9.145,12.275,15.405,...........                      

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…