Third term of a series in Ap is 8 and 7th term is 25.5 find 23rd term
Answers
Answer:
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,
⇒T
3
=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,
⇒T
9
=3T
3
+2=(3×8)+2=24+2=26
But
⇒T
9
=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=
6
18
=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:
⇒S
n
=
2
n
[2a+(n−1)d]
⇒Substitute n=19, a=2 and d=3 in S
n
=
2
n
[2a+(n−1)d] as follows:
⇒S
19
=
2
19
[(2×2)+(19−1)3]=
2
19
[4+(18×3)]=
2
19
(4+54)=
2
19
×58=19×29=551
Hence, the sum of the first 19 terms of an A.P is S
19
=551.
Answer:
Step-by-step explanation: