# 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?

Devesh Lall
35 Points
12 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

Devesh Lall
35 Points
12 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

Devesh Lall
35 Points
12 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