A man saves Rs 32,000 during first year Rs 36,000 in the next year and Rs 40,000 in the third year. if he continues his savings in this sequence in how many years will he save Rs 2,00,000.
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Answered by
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The savings are increased in arithmetic sequence.
The nth term of arithmetic sequence
= a1 + (n-1)*d
200000= 32000 + (n-1)*4000
200000 - 32000 = (n-1)*4000
168000 = (n-1) * 4000
168000/4000 = (n-1)
n-1 = 42
n = 42+1
= 43
In the 43rd year the savings will become 2,00,000 rupees
The nth term of arithmetic sequence
= a1 + (n-1)*d
200000= 32000 + (n-1)*4000
200000 - 32000 = (n-1)*4000
168000 = (n-1) * 4000
168000/4000 = (n-1)
n-1 = 42
n = 42+1
= 43
In the 43rd year the savings will become 2,00,000 rupees
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