    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 $=₹\left(\frac{2}{3}×3600\right)$ $=₹\left(2×1200\right)\phantom{\rule{0ex}{0ex}}=₹2400$Let $a$ be the first instalment and $d$ be the common difference between them.Step $\mathbf{1}$ : Find a using Sn for $\mathbf{30}$ instalments.We know from the formula of AP that,${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$Substituting the values of $a$ and $d$,${S}_{30}=\frac{30}{2}\left[2a+\left(30-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒2400=15\left[2a+29d\right]\phantom{\rule{0ex}{0ex}}⇒\left[2a+29d\right]=\frac{2400}{15}\phantom{\rule{0ex}{0ex}}⇒\left[2a+29d\right]=160\phantom{\rule{0ex}{0ex}}⇒2a=160-29d\phantom{\rule{0ex}{0ex}}⇒a=\frac{160-29d}{2}...............\left[1\right]$Step $\mathbf{2}$ : Find a using Sn for $\mathbf{40}$ instalments.Similarly, $a$ and $d$ are calculated for $40$ instalments${S}_{40}=\frac{40}{2}\left[2a+\left(40-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒3600=20\left[2a+39d\right]\phantom{\rule{0ex}{0ex}}⇒\left[2a+39d\right]=180\phantom{\rule{0ex}{0ex}}⇒2a=180-39d\phantom{\rule{0ex}{0ex}}⇒a=\frac{180-39d}{2}.............\left[2\right]$Step 3 : Find a and d using above steps.Subtracting $eq.\left[1\right]$ from $eq.\left[2\right]$, we get$a-a=\left(\frac{180-39d}{2}\right)-\left(\frac{160-29d}{2}\right)\phantom{\rule{0ex}{0ex}}⇒0=\frac{180-39d-160+29d}{2}\phantom{\rule{0ex}{0ex}}⇒0=\frac{20-10d}{2}\phantom{\rule{0ex}{0ex}}⇒20-10d=0\phantom{\rule{0ex}{0ex}}⇒10d=20\phantom{\rule{0ex}{0ex}}⇒d=\frac{20}{10}\phantom{\rule{0ex}{0ex}}⇒d=2$Substituting the value of $d$,$a=\frac{160-29×\left(2\right)}{2}\phantom{\rule{0ex}{0ex}}⇒a=\frac{160-58}{2}\phantom{\rule{0ex}{0ex}}⇒a=\frac{102}{2}\phantom{\rule{0ex}{0ex}}⇒a=51$Hence, $\mathbf{₹}\mathbf{51}$ was paid in the first instalment.  Suggest Corrections  0      Similar questions  Explore more