CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A man arranges too pay off a debt of 3600 by 40 annual instalments which form an arithmetic series.

When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.


Open in App
Solution

Total amount of debt to be paid in 40 instalments =3600

After 30 instalments, one-third of his debt is unpaid which means that two-third of his payment is done.

Therefore, the amount he paid in 30 instalments =(23×3600)

=(2×1200)=2400

Let a be the first instalment and d be the common difference between them.

Step 1 : Find a using Sn for 30 instalments.

We know from the formula of AP that,

Sn=n2[2a+(n-1)d]

Substituting the values of a and d,

S30=302[2a+(30-1)d]2400=15[2a+29d][2a+29d]=240015[2a+29d]=1602a=160-29da=160-29d2...............[1]

Step 2 : Find a using Sn for 40 instalments.

Similarly, a and d are calculated for 40 instalments

S40=402[2a+(40-1)d]3600=20[2a+39d][2a+39d]=1802a=180-39da=180-39d2.............[2]

Step 3 : Find a and d using above steps.

Subtracting eq.[1] from eq.[2], we get

a-a=(180-39d2)-(160-29d2)0=180-39d-160+29d20=20-10d220-10d=010d=20d=2010d=2

Substituting the value of d,

a=160-29×(2)2a=160-582a=1022a=51

Hence, 51 was paid in the first instalment.


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image