3. The sum of first 13 terms and the sum of next 12 terms of an arithmetic sequence are
equal. If its common difference is 3,
a) How many times of the common difference will be the difference between 14th
and first terms of this sequence ?
b) What is the 13th term of this sequence ?
c) What is the sum of first 25 terms of this sequence ?
Answers
Answer:
a answer
Step-by-step explanation:
this is the answer
Answer:
Step-by-step explanation:
Sum of n terms of AP is given by Sₙ = n/2 [2a + (n-1)d]
nth term of AP is given by tₙ = a + (n - 1)d
Let a be the first term and d be the common difference.
Given,
d = 3
Sum of first 13 terms = sum of next 12 terms.
Sum of first 13 terms of AP = 13/2 [ 2a + (13 - 1)3] = 13/2( 2a + 36) -- (1)
Now to calculate the next 12 terms, we need to the first term of this this series.
First term of next 12 terms = 14th term of series
=> t₁₄ = a + (14 - 1)d = a + 13d = a + 39
Sum of next 12 terms = 12/2[2(a + 39) + (12 - 1)3] = 12/2( 2a + 111) --- (2)
=> 13/2(2a + 36) = 12/2(2a + 111)
=> 26a + 468 = 24a + 1332
=> 2a = 864
=> a = 432.
a) t₁₄ - t₁ = [(a+13d) - a] = 13d = 13 * 3 = 39
b) t₁₃ = a + 12d = 432 + 12 * 3 = 468
c) S₂₅ = 25/2 [ 2 * 432 + (25 - 1)3] = 25/2[864 + 72] = 25 * 468 = 11,700.