Question

A sum of 1000 is invested at a interest of 8 % simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.

Answer 1

$\huge\tt\red{SI=\frac{P×R×T}{100}}$

$\small\sf\underline\pink{So,the\:interest\:at\:the\:end\:of\:the\:1st\:year=}$

$\longrightarrow$$\large\tt\purple{ \frac{Rs.1000 \times 8 \times 1}{100} = Rs.80}$

$\small\sf\underline\pink{The\:interest\:at\:the\:end\:of\:the\:2nd\:year=}$

$\longrightarrow$$\large\tt\purple{ \frac{Rs.1000 \times 8 \times 2}{100} = Rs.160}$

$\small\sf\underline\pink{The\:interest\:at\:the\:end\:of\:the\:3rd\:year=}$

$\longrightarrow$$\large\tt\purple{\frac{Rs.1000 \times 8 \times 3}{100} = Rs. 240}$

$\small\sf\underline\blue{So,the\:interest(in\:Rs.)\:at\:the\:end\:of\:the\:1st,2nd}$$\small\sf\underline\blue{and\:3rd\:years\:respectively\:are\:80,160\:and\:240,....}$

$\small\sf\underline\green{It \:is \:an\:AP\: as\: the\: difference\: between\: the}$

$\small\sf\underline\green{consecutive\: terms\: in\: the \:list \:is\:80}$

$\small\sf\underline\orange{So, to \:find \:the\: interest\: at\: the\: end \:of \:30\: years \:we}$

$\small\sf\underline\orange{shall\: find\:a30}$

$\longrightarrow$$\large\tt\gray{a30=a+(30-1)d}$

$\longrightarrow$$\large\tt\gray{80+29×80}$

$\longrightarrow$$\large\tt\gray{2400}$

$\therefore$$\small\sf\underline\red{So,the\:interest\:at\:the\:end\:of\:30\:years\:will\:be}$

$\small\sf\underline\red{Rs.2400}$