An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the 25 terms
Answers
Correct question:-
Q// An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the sum of 25 terms?
Required Answer:-
Given:
- Total number of terms = 50
- Third term (a3) = 12
- Last term (a50) = 106
Now,
By nth term formula,
Where, an, a, n and d are the last term, first term, term number and common difference respectively.
Then,
- a + 2d = 12 ------(1)
- a + 49d = 106 ------(2)
Subtracting (1) from (2),
➛ a + 49d - (a + 2d) = 106 - 12
➛ a + 49d - a - 2d = 94
➛ 47d = 94
➛ d = 2
Then, a = 12 - 4 = 8
To Find:
- Sum of 25 terms of the AP.
- Then, n = 25, a = 8, d = 2.
By using formula,
➛ S25 = 25/2 {2(8) + (25-1)2}
➛ S25 = 25/2 {16 + 24 × 2}
➛ S25 = 25/2 {64}
➛ S25 = 25 × 32
➛ S25 = 800
Hence,
The required value of the sum of 25 terms of the above AP is 800 (Ans).
Answer:
Correct question:-
Q// An AP consists of 50 term of which 3rd term is 12 and the last term is 106 find the sum of 25 terms?
Required Answer:-
Given:
Total number of terms = 50
Third term (a3) = 12
Last term (a50) = 106
Now,
By nth term formula,
\boxed{\rm{\large{a_n = a + (n - 1)d}}}
a
n
=a+(n−1)d
Where, an, a, n and d are the last term, first term, term number and common difference respectively.
Then,
a + 2d = 12 ------(1)
a + 49d = 106 ------(2)
Subtracting (1) from (2),
➛ a + 49d - (a + 2d) = 106 - 12
➛ a + 49d - a - 2d = 94
➛ 47d = 94
➛ d = 2
Then, a = 12 - 4 = 8
To Find:
Sum of 25 terms of the AP.
Then, n = 25, a = 8, d = 2.
By using formula,
\boxed{\rm{\large{S_n = \dfrac{n}{2} \bigg (2a + (n - 1)d \bigg)}}}
S
n
=
2
n
(2a+(n−1)d)
➛ S25 = 25/2 {2(8) + (25-1)2}
➛ S25 = 25/2 {16 + 24 × 2}
➛ S25 = 25/2 {64}
➛ S25 = 25 × 32
➛ S25 = 800
Hence,
The required value of the sum of 25 terms of the above AP is 800 (Ans).