The sum of the first five terms of an AP and the sum of the first seven terms of the
same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of
its first twenty terms.

ans is 970

ans is 970

Answer is 470...!!

A lill mistake at las....

Are u a gal??

S₅+S₇= n/2 (2a +4d ) + n/2 (2a+6d)

  = 5(a+2d)+7(a+3d)

  = 5a + 10d+7a+21d

  = 12a + 31d

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

  S₁₀= 10/2 (2a+9d)

235 = 5(2a+9d)

2a+9d=47 ......2   ;multiplying equation 2 from 6

12a + 54d=282  ......3

from 1&3

54d-31d=282-167

23d=115

d = 2  ;  putting the value of d in equation 2

2a + 18 = 47

a=29/2

therefore, S₂₀=20/2(2*29/2+19*2)

  = 10 (29+ 38)

  = 10*67

  = 670 ........ans

Nopes its 470....!!

how you got it

Buddy i took this question in other way..

I thought S5 = S7 = 167 ... lol

i have made a mistake d = 5

2a + 45 =47

a=1

S₂₀=10(2+95)

  =  10*97

  =970............ans

thank you

then thumbs up please!

yes, does that make any difference in the ans

 let the sum of first five terms =s5 and seven terms =s7
the formula for sum of first n terms = sn=n/2[2a+(n-1)d] n= number of terms, a is the first term and d is the difference
s5+s7=167= 5/2[2a+4d]+ 7/2[ 2a+ 6d] =167
5[a+2d]+7[a+3d]=167
5a+10d+7a+21d=167
12a+31d=167


235=10/2[2a+9d]
235=5[2a+9d]
235=10a+45d
2a+9d=47(dividing by 5)
solving the two equations we get d=5 a=1 
sum of first twenty terms = 20/2[2+19*5]
10[2+95]
970

in an a.p first term a=2 and the sum of the first five terms is equal to one fourth of the sum of the next five terms show that T20=12........plz do ans fa dis query itz urgent