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Question

The third term of an A.P. is 8 and the ninth term of the A.P. exceeds three times the third term by 2. Find the sum of its first 19 terms.

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Solution

We know that the nth term of an A.P with first term a and common difference d is Tn=a+(n1)d.

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

T3=a+(31)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+(91)d=a+8d, thus,

a+8d=26.........(2)

Now, subtract equation 1 from equation 2 as follows:

(aa)+(8d2d)=268

6d=18
d=186=3

Substitute d=3 in equation 1:

a+(2×3)=8a+6=8a=86=2

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

Sn=n2[2a+(n1)d]

Substitute n=19, a=2 and d=3 in Sn=n2[2a+(n1)d] as follows:

S19=192[(2×2)+(191)3]=192[4+(18×3)]=192(4+54)=192×58=19×29=551

Hence, the sum of the first 19 terms of an A.P is S19=551.

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