1. The sum of first and 21st terms of an arithmetic sequence is 140.
a. What is the 11th term?
b. Write the sequence, if first term is 10.
c. Calculate the sum of first 11 terms of this sequence.
d. Find the sum of first 11 terms of the sequence 20,25,30..
Answers
Let the first term be 'a' and common difference be 'd'. Given, sum of 1st term and 21st term is 140.
=> 1st term + 21st term = 140
=> a + (a + (21 - 1)d) = 140
=> 2a + 20d = 140
=> 2(a + 10d) = 140
=> a + (11 - 1)d = 140/2
=> 11th term = 70
a) 11th term of the given series is 70.
b) If a = 10
=> 11th term = 70
=> 10 + 10d = 70
=> d = 6 thus, series is
a, a + d, a + 2d, a + 3d...
10, 10+6, 10+2(6), 10+3(6)...
10, 16, 22, 28....
c) sum of n term = (n/2)(a + l)
=> sum = (11/2) (10 + 70)
=> sum = 11/2 × 80 = 440
d) sum of n term = (n/2)(2a + (n-1)d)
Here, a = 20, d = 25 - 20 = 5, n = 11
S = (11/2) [2(20) + (11 - 1)5]
S = (11/2) [90]
S = 495
Solved using :
nth term = a + (n - 1)d
sum = (n/2) (a + l) = (n/2)(2a + (n - 1)d)
*l is last or nth term
Required Answer
a]
We know that
a + [a + (21 - 1)d] = 140
a + a + (21 - 1) d = 140
2a + 20d = 140
2(a + 10d) = 140
a + 10d = 140/2
a + 10d = 70
11 the term can be written as a + 10d
b]
Series = a, a + d, a+ 2d
Series = 10, 16, 22
c]
Sum of 11th term = n/2 × (a + l)
Sum of 11th term = 11/2 × (10 + 70)
Sum of 11th term = 5.5 × 80
Sum of 11th term = 440
d]
Sum of 11 terms
Sₙ = 11/2{2 × 20 + (11 - 1)5}
S₁₁ = 11/2{40 + 10(5)}
S₁₁ = 11/2{90}
S₁₁ = 495