Find the sum of the first 30 natural numbers which leave the remainder 2 when divided by 5.
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Answer:
676
Step-by-step explanation:
The first number is 10 which when divided by 7 leaves a remainder of 3.
The last number is 94.
The sum is an AP whose first term is 17 and the last one is 94.
Tn = a+(n-1)d or
94 = 10 + (n-1)*7, or
84 = (n-1)*7, or
n-1 = 12, or
n = 13.
S12 = (13/2)[2*10+(13–1)*7]
= (13/2)[20+84]
= (13/2)[104]
= 676.
Answer : 676.
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