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# 1. Find the sum of the 1st 25 terms on A.P. whose nth term is given by tn=2-3n.2. Find the sum of all two digits natural numbers which when divided by 3 yield 1 as remainder.3. A man borrows Rs. 1000 and agrees to repay with a total interest of Rs. 140 in 12 installments. Each instalment being less than the preceding by Rs. 10. What should be his 1st instalment?

## 3 Answers

9 years ago

SOLUTION FOR QUESTION NO.-1 --------  tn=2-3n (given)

1st term=t1=2-3x1

=-1

sly.

2nd term=t2=2-3x2

=-4

now, d(common difference)=t2-t1=-4-(-1)=-3

since n=25

therefore, sum of 1st 25 terms=sn=n/2[2a+(n-1)d]

=25/2[2x-1+(25-1)-3]

=-925

9 years ago

SOLUTION FOR QUESTION NO.-2 ---------

since 2-digits no.s starts from 10 & ends at 99

therefore required no.s which when divided by 3 leaves remainder 1 are:- 10,13,16,19,................,94,97

here the 1st term(a) is 10 & c.d.(d) is 3 & the last term is 97(l).

now,

l=a+(n-1)d

97=10+(n-1)3

n=30

sn=n/2[2a+(n-1)d] (sum of the terms to nth term)

=30/2[2x10+29x3]

=1605

9 years ago

SOLUTION FOR QUESTION NO.-3

Since amt borrowed by a man earlier = Rs 1000

"      total interst offerd to him in 12 installments = Rs 140

Now, let total overall amt be with the man be Sn =Rs (1000+140)

= Rs 1140

"      "   his 1st installment be a

"      " n be the no of installments = 12

"      " d " =-10

"  using the formula :-   Sn=n/2[2a+(n-1)d]

1140=12/2[2a+11x-10]

1140=6[2a-110]

1140=12a-660

12a=1140+660

12a=1800

a= Rs 150

Therefore his required 1st installment = Rs 150

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