if 3+5+7+9...........to n terms/ 5+8+11...to 10 terms = 7

then find the value of n

 a =3

d = 5-3 =2

a_n = a+ (n-1)d 

      = 3+ 2n-2 = 2n+1 

S_n = n/2 (a+ a_n)

     = n/2 (3+2n+1)

    = n(2+n)

   = n²+2n ........(1)

for 2nd,

a =5

d = 8-5 = 3

a₁₀ = a+ 9d  = 5+ 27 = 32

S₁₀ = 10/2 (a+an)

       = 5 (5+32)

       = 185  ..........(2)

equ. 1 by equ. 2 = 7

n² +2n / 185 =7

n² +2n = 1295

n² +2n -1295 =0

now by splitting the middle term,

n² + (37 -35)n -1295 =0

n² +37n -35n -1295 =0

n(n+37) -35(n+37) 

(n+37) (n-35)

=> n =-37 and n =35

since , n cannot be negative . therefore n =35