the sum of first 7 terms in a series of 10 terms in an ap is 28, the sum of the last 7 terms is 49. find the sum of all the 10 terms
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Answer:
Sum of 10 terms of an AP = 55
Step-by-step explanation:
Given:
- There are ten terms in an AP
- Sum of first 7 terms (1-7) = 28
- Sum of last 7 terms (4-10) = 49
To find:
- Sum of 10 terms of an AP
Required Formulae:
1. Term Formula : nth term of an AP
a(n) = a + (n - 1)d
where,
- a = first term of an AP
- d = common difference
2. Sum Formula : sum of n terms of an AP
S(n) = n/2 * ( 2a + (n - 1)d )
where,
- a = first term of an AP
- d = common difference
Solution:
Sum of first 7 terms = 28
Here, first term = a(1) & common difference = d
=> S = 7/2 * ( 2a + (7 - 1)d ) = 28
=> 7/2 * ( 2a + 6d ) = 28
=> 1/2 * 2 ( a + 3d ) = 4
=> a + 3d = 4 (Eq 1)
Sum of last 7 terms = 49
Here, first term = a(4) & common difference = d
Using Term formula, we get
a(4) = a + (4 - 1)d = a + 3d
But a + 3d = 4 [From Eq 1]
=> a(4) = 4
S = 7/2 * ( 2a(4) + (7 - 1)d ) = 49
=> 7/2 * ( 2(4) + 6d ) = 49
=> 1/2 * ( 8 + 6d ) = 7
=> 8 + 6d = 14
=> 6d = 14 - 8 = 6
=> d = 1
Putting d = 1 in Eq 1, we get
a + 3(1) = 4
=> a + 3 = 4
=> a = 1
Sum of 10 terms of an AP [ S(10) ] = 10/2 * ( 2a + (10 - 1)d )
=> 5 * ( 2(1) + 9(1) )
=> 5 * ( 2 + 9 )
=> 5 * 11 = 55
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