The sum of first 7 terms of an AP is 63 and the sum of its next 7 terms is 161 then find the 28th term of the AP.
Answers
Solution:
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
Answer:
Step-by-step explanation:
Solution :-
Since, S(n) = n/2[2a + (n - 1)d]
Given, S(7) = 63
Hence, S(7) = 7/2[2a + 6d] = 63
or, 2a + 6d = 18 .... (i)
Now, the sum of 14 terms is
S(14) = S(first 7) + S(next 7)
= 63 + 161
= 224
Then, 14/2[2a + 13d] = 32 ...(ii)
On Subtracting Eq (i) and (ii), we get
⇒ (2a + 13d) - (2a + 6d) = 32 - 18
⇒ 7d = 14
⇒ d = 14/7
⇒ d = 2
Purring d's value in Eq (i), we get
⇒ 2a + 6d = 18
⇒ 2a + 6(2) = 18
⇒ 2a + 12 = 18
⇒ 2a = 18 - 12
⇒ 2a = 6
⇒ a = 6/2
⇒ a = 3
Since, a(n) = a + (n - 1)d
⇒ a(28) = 3 + 2 × (27)
⇒ a(28) = 57
Hence, the 28th term of the AP is 57.