Question

Person borrows 4500Rs. and promises to pay back (without any interest ) in 30 instalments each of values Rs.10 more than last (preceding one ) find first and last instalments

Answer 1

first installment = x
second installment = X + 10
third installment = X + 20 and so on.

consider an A.P
that is,
x, x + 10, X + 30, ..... and so on.

total number of terms = 30
whose,
common difference is (X+10)-x
d = 10

person have to pay 4500. (which is sum of all the 30 terms)

by formula,
we have,

$sum \: of \\ n \: terms = \frac{n}{2} (2a + (n - 1)d) \\ sum \: of \\ 30 \: terms = \frac{30}{2} (2x + 29d)$
we know that common difference is 10
we have also given that sum of 30 terms is 4500.
So,

$4500= \frac{30}{2} (2x + 29d) \\ 4500= 15(2x + 290) \\300 = 2x + 290 \\ 2x = 10 \\ x = 5$
first stallment = 5

We also know to derive nth term of an A.P
by Formula,

$nth \: term \: of \\ an \: ap = a + (n - 1)d \\ 30th \: term = 5 + 290 \\ 30th \: term = 295$
his last installment will be rupees 295.

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