Question

the 4th term of an ap is 11 and the 8th term exceeds twice the fourth term by 5. find the sum of the first 30 term.
solution​

Answer 1

Explanation:

4

th

term=11

∴ a+3d=11

⇒8

th

term=2(4

th

term)+5

⇒a+7d=27

⇒(a+7d)−(a−3d)=27−11

⇒4d=16

⇒d=4

⇒a=−1

Formula: S

n

=

2

n

[2a+(n−1)d]

Sum of first 50 terms =

2

50

(2(−1)+49(4))

=25(194)

=4850

Answer 2

Answer:

Here's your answer

Explanation:

4th term = 11

Therefore, a + 3d = 11

= 8th term = 2 (4th term) + 5

= a +7d = 27

= (a + 7d) - (a - 3d) = 27 - 11

= 4d = 16

= d = 4

= a = 1

Formula :- Sn = n/2 [2a + (n - 1) d]

Sum of first 30 terms = 30/2 (2(-1) + 49 (4)

= 30/2 × 2 - 1 + 49 × 4

= 15 × (-2) + 196

= 15 × 194

= 2910

Therefore, 2910 is the correct answer ✓✓

Previous bro's answer is incorrect