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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?
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
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
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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