A man borrows Rs. 12,000 and promises to pay back in 20 instalments, each of value Rs 30 more than the last. Find the first instalment.

-
Q. A man borrows Rs. 12,000 and promises to pay back in 20 instalments, each of value Rs 30 more than the last. Find the first instalment.
Solution:

Let us assume that a=first installment, d=common difference, n= number of installments.
Here, Sₙ=12,000, n=20, d=30.
To find a=?













Therefore the first installment is Rs. 315

Comments

Post a Comment

Popular posts from this blog

The sum of three consecutive terms in A.P. is 24 product is 440, find the terms.

-
Q. The sum of three consecutive terms in A.P. is 24 product is 440, find the terms. Solution: Let us assume that the numbers are a-d, a, a+d Given, (a-d)+a+(a+d)=24 ⇒3a=24 ⇒a=24/3 ⇒a=8 and (a-d)*a*(a+d)=440 ⇒(8-d)*8*(8+d)=440 ⇒(8-d)(8+d)=440/8 ⇒8²-d²=55   (since (a+b)(a-b)=a²-b²) ⇒64-d²=55 ⇒-d²=55-64 ⇒-d²=-9 ⇒d²=9 ⇒d²=3² ⇒d=±3 If d=3, then the numbers are 8-3,8,8+3 => 5,8,11 If d=-3, then the numbers are 8-(-3), 8, 8+(-3)=>8+3, 8, 8-3 =>11, 8, 5
Read more

The third and sixth terms of a series in A.P are 13 and 31 respectively. Find 20th term.

-
Q. The third and sixth terms of a series in A.P  are 13 and 31 respectively. Find 20th term. Solution: Let us assume that first term=a and common difference=d. Given, third term=t₃=13 ⇒a+(3-1)*d=13 ⇒a+2d=13...................(i) and sixth term=t₆=31 ⇒a+(6-1)*d=31 ⇒a+5d=31...............(ii) (ii)-(i)⇒(a+5d)-(a+2d)=31-13 ⇒a+5d-a-2d=18 ⇒a-a+5d-2d=18 ⇒0+3d=18 ⇒3d=18 ⇒d=18÷3 ⇒d=6 Substituting the value of d=6 to the equation (i),we get (i)⇒a+2*6=13 ⇒a+12=13 ⇒a=13-12 ⇒a=1 20th term=t₂₀=a+(20-1)*d =1+19*6 =1+114 =115.
Read more