how many terms of an AP must be taken for their sum to be equal to 120 if its third term is 9 and the difference between the seventh and second term is 20

 Let the first term is a, and the common difference is d of the AP.
Tnth = a + (n-1)d
T3rd = 9 = a + (3-1)d
⇒a+2d = 9 (i)
and T7th-T2d = {a +(7-1)d }-{a+(2-1)d} = 20
⇒ {a+6d}-{a+d}=20
⇒5d = 20
⇒d =20/5 = 4
Substituting the value of d in equation (i), we get a + 2тип4 = 9, a = 1.

Sum upto nth term is given by Snth = n/2{2a+(n-1)d}
So Snth = 120 = n/2{2(1) + (n-1)4}
⇒ n{2+4n-4} = 240
⇒n{4n-2} = 240
4n2-2n-240 = 0
⇒ 2n2 -n -120 =0
⇒2n2 -16n+15n -120 = 0
⇒ 2n(n-8)+15(n-8) = 0
⇒ (2n+15)(n-8) = 0
n = -15/2, 8 so taking n = 8, as n can never be fraction or negative.
So sum of 8 terms of the AP =120.

  • 3

given , 3rd term = 9

7th term - 2nd term = 20

ie, a+2d = 9 ..............1

(a+6d) - (a+d) = 20

a+6d-a-d = 20

5d = 20

d = 4

substituting the value of d in eqn 1

a+2d = 9

a+2(4) = 9

a+8 = 9

a = 9-8

a = 1

given Sn = 120

now , Sn = n/2 [2(1)+(n-1) (4)]

= n/2 [2+4n-4]

n/2 x2 [1+2n-2]

2 and 2 get's cancel here i have taken 2 as common.

= n [1+2n-2] = 120

= n+2n2 2n = 120

= 2n2-n-120 = 0

splitting the middle term.......

= 2n2-16n+15n-120 = 0

= 2n[n-16] +15[n-16] = 0

[n-16] = 0 and [2n+15] = 0

n= 16 n = -15/2 (it is an extraneous solution)

therefore 16 terms should be taken to make the sum of 120.

  • -2
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