Sum of n Terms of an Arithmetic Progression :
To find the sum of the series, we use the formula given below.
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
To find the first term, common difference and number of term, we use the formula given below.
an = a + (n - 1)d
Question 1 :
If the sum of 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms.
Solution :
S7 = 49
S17 = 289
Sn = (n/2) [2a + (n - 1) d]
S7 = (7/2) [2 a + (7 - 1) d]
49 = (7/2) [ 2a + 6d]
(49 x 2)/7 = 2 a + 6 d
14 = 2 a + 6 d
a + 3 d = 7 -----(1)
S17 = (17/2) [2 a + (17 - 1) d]
289 = (17/2) [ 2a + 16d]
(289 x 2)/17 = 2 a + 16 d
34 = 2 a + 16 d
a + 8 d = 17 -----(1)
(1) - (2)
a + 3 d = 7
a + 8 d = 17
(-) (-) (-)
-------------
- 5 d = -10
d = 2
By applying the value of d in (1), we get
a + 3 (2) = 7
a + 6 = 7
a = 7 - 6
a = 1
To find the sum of first n terms, we have to apply the values of a and in the Sn formula
Sn = (n/2) [2a + (n - 1) d]
Sn = (n/2) [2(1) + (n - 1) (2)]
= (n/2) [2 + 2 n - 2]
= (n/2) [2 n]
= n²
Question 2 :
Show that a₁, a₂,............ an form an AP where an is defined as below
(i) an = 3 + 4 n
(ii) an = 9 - 5 n
Also find the sum of 15 terms in each case.
Solution :
(i) a n = 3 + 4 n
n = 1 a₁ = 3 + 4(1) = 7 |
n = 2 a₂ = 3 + 4(2) = 11 d = a₂ - a₁ = 11 - 7 d = 4 |
So the series will be in the form 7 + 11 + ...........
Now we need to find sum of 15 terms
S15 = (n/2) [ 2 a + (n - 1) d]
= (15/2) [2(7) + (15-1) 4]
= (15/2) [14 + 14(4)]
= (15/2) [14 + 56]
= (15/2) [70]
= (15 x 35)
S15 = 525
(ii) an = 9 - 5 n
Solution :
n = 1 a1 = 9 - 5(1) = 4 |
n = 2 a2 = 9 - 5(2) = -1 d = a2 - a1 = -1 - 4 = -5 |
So the series will be in the form 4 + (-1) + ...........
Now we need to find sum of 15 terms
S15 = (n/2) [ 2 a + (n - 1) d]
= (15/2) [2(4) + (15-1) (-5)]
= (15/2) [8 + 14(-5)]
= (15/2) [8 - 70]
= (15/2) [-62]
= 15 x (-31)
= -465
After having gone through the stuff given above, we hope that the students would have understood, sum of n terms of an arithmetic progression.
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