Question

The 3rd team of an AP
IS 8and
the 9th
term of an AP Exceeds three
times third term by a find the
Sum of its first 25 terms​

Answer 1

Answer:

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Step-by-step explanation:

We know that the nth term of an A.P with first term a and common difference d is T

n

=a+(n−1)d.

Here, it is given that the third term of an A.P is 8, therefore,

⇒T3 =a+(3−1)d

⇒8=a+2d

⇒a+2d=8..(1)

It is also given that the ninth term of an A.P exceeds three times the third term by 2, therefore,

⇒T9=3T3+2

=(3×8)+2

=24+2

=26

But

⇒T9=a+(9−1)d=a+8d,

thus,

⇒a+8d=26..(2)

Now, subtract equation 1 from equation 2 as follows:

⇒(a−a)+(8d−2d)=26−8

⇒6d=18

⇒d= 618

=3

Substitute d=3 in equation 1:

a+(2×3)=8

⇒a+6=8

⇒a=8−6=2

We also know that the sum of n terms of an A.P with first term a and common difference d is:

⇒Sn= 2n[2a+(n−1)d]

⇒Substitute n=19, a=2 and d=3 in

Sn= 2n[2a+(n−1)d] as follows:

⇒S19 = 219 [(2×2)+(19−1)3]

=219 [4+(18×3)]

=219(4+54)

= 219×58

=19×29

=551

Hence, the sum of the first 19 terms of an A.P is S

19

=551.