7. Find the sum of (a) the first six even numbers, (b) the first n even numbers. For what value of n does the sum in (b) equal 90? For what value of n, does the sum first exceed 1000? The third term in an arithmetic series is 9 and the eleventh term is 25.
Answers
Answer is in attachment. Once check.
Given : the first six even numbers,
To Find : Sum of first six even numbers
Sum of the first n even numbers.
Value of for which the sum in (b) equal 90
Solution:
even numbers
2 , 4 , 6
AP a = 2 , d = 2
Sum of 6 even numbers = (6/2) ( 2 * 2 + 5 * 2)
= 3 ( 4 + 10)
= 42
Sum of the first n even numbers.
= (n/2) (2 * 2 + (n - 1 ) 2)
= (n/2) ( 2n + 2)
= n ( n + 1)
n ( n + 1) = 90
=> n = 9 as 9 * 10 = 90
the sum first exceed 1000
n ( n + 1) > 1000
31 * 32 = 992
32 * 33 = 1056
Hence for n = 32 sum first exceed 1000
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