Question

4) Vijay invested certain amount each year in a bank. He started with Rs. 500 in the first
year and invested Rs. 200 more each year than the previous year's investment. What
will be his investment after 12 years.​

Answer 1

AnswEr :

• Vijay Started with Rs. 500
• Invested Rs. 200 more each year than previous year.
• Find the Investment after 12 years.

$\begin{aligned}\textsf{First Year, Second Year, Third Year...so on}\\\textsf{Rs. 500, Rs. (500+200),Rs. (500+200+200)...so on}\\\textsf{Rs. 500, Rs. 700, Rs. 900...so on}\\\textbf{\textdagger}\:\boxed{\texttt{This is in Arithmetic Progression}}\end{alinged}$

$\bf{where}\begin{cases}\textsf{First Term (a) = Rs. 500}\\\textsf{Common Difference (d) = Rs. 200}\\ \textsf{No. of Terms (n) = 12 years}\\\textsf{Find the Sum of Terms?} \end{cases}$

$\rule{100}{2}$

For any Arithmetic Progression ( AP ), the sum of n terms is Given by :

$\textbf{\textdagger}\quad\large\boxed{\sf S_n = \dfrac{n}{2}\bigg(a + l\bigg)}$

where,

• n = no. of terms
• a = First Term
• l = Last Term

Since, the last term is also the nth term :

◗ l = a + (n – 1)d

• Substituting for l in the formula for sum :

$\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(a + [a + (n -1)d]\bigg)\\ \\\longrightarrow \tt S_n = \dfrac{n}{2}\bigg(2a+(n-1)d\bigg) \\ \\\longrightarrow \tt S_n = \cancel\dfrac{12}{2}\bigg((2 \times Rs.500)+(12-1) \times Rs.200\bigg) \\ \\\longrightarrow \tt S_n = 6(Rs.1000+(11 \times Rs.200)) \\ \\\longrightarrow \tt S_n =6(Rs.1000 + Rs.2200) \\ \\\longrightarrow \tt S_n =6\times Rs.3200 \\ \\\longrightarrow \large\boxed{ \red{\tt S_n =Rs. \:19200}}$

⠀

∴ Therefore, Total Investment of Vijay after 12 years will be Rs. 19200.

Answer 2

❏ QuesTion :-

→ Given above ↑

❏ Answer :-

→ Total investment after 12 years is ₹19200

❏ To Find :-

→ His total investment after 12 years .

❏ SoluTion :-

According to the question -

He invested ₹ 500 in first year ,

after that he added ₹ 200 per year investment

Hence ,

sequence of his investment every year is

→500 , (500 + 200 ) , (500 + 200 ),....

The sequence of his investment is an arithmetic progression ,

hence , according to the condition -

$\to \sf \red{ first \: term \: (a)} = \: 500 \\ \\ \sf \to \green{common \: difference \: } (d) =\orange{( 500 + 200) - 500} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{ (d) = 200} \\ \to \red{\sf he \: invested \: till \: 12 \: years }\: \\ \\ \: \: \: \star \: \sf hence \: \: n = 12 \\ \\ \bf \: we \: know \: that \\ \\ \to\green{ \sf \: S_n \: }=\orange{ \sf \frac{n}{2} \{ \: 2a + (n - 1)d \}} \\ \\ \to \sf S_{12} = \blue{ \frac{12}{2} \{2 \times 500 + (12 - 1) \: 200 \} }\\ \\ \to \sf S_{12} = 6 \{1000 + 11 \times 200 \} \\ \\ \to\red{ \sf S_{12} = 6 \{1000 + 2200 \} }\\ \\ \to \sf \: S_{12} = 6 \times 3200 \\ \\ \boxed{ \to \sf \: S_{12} = 19200}$

His total investment after 12 years is ₹ 19200

\__________|__________/

Hope it will help you .