The sum of first terms of an A. p is 63.
The sum of the next seven terms
exceeds the sum of first seven
terms by 28
Answers
Answer:
Sₙ = (n/2) [2a + (n -1)d]
Sum of the first 7 terms of an A.P is 63 i.e. S₇ = 63
(7/2) [ 2a + 6d ] = 63
2a + 6d = (63 * 2)/7
2a + 6d = 18 [Equation 1]
Sum of its next 7 terms = 161 [Given in Question]
Sum of first 14 terms = Sum of first 7 terms + Sum of next 7 terms.
S₁₄ = 63 + 161 = 224
And,
Applying formula Sₙ = (n/2) [2a + (n -1)d] on 14 terms we get,
(14/2) [2a + 13d] = 224
7[2a + 13d ] = 224
2a + 13d = 32 [Equation 2]
By Subtracting Equation 1 from Equation 2,
[2a + 13d] - [2a + 6d] = 32 - 18
2a + 13d - 2a - 6d = 14
7d = 14
d = 2
Putting value of d in Equation 1 to obtain a,
2a + 6d = 18
2a + 6(2) = 18
2a + 12 = 18
2a = 18 - 12
2a = 6
a = 6/2
a = 3
Now, we have to find the 28th term of AP and we already know first term of AP i.e. a = 3 and common difference of AP i.e. d = 2
Let's apply the formula,
aₙ = a + (n-1)d
a₂₈ = a + (28 - 1)d
a₂₈ = 3 + (28 - 1)2
a₂₈ = 57
The 28th term of AP is 57
Step-by-step explanation: