Question

Sachin invested in a national saving in the first year he invested rupees 500 in the second year rupees 900 and so on find the total amount that he invested i 22 years​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Total\:amount=1,03,400\:rupees}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{Given : }} \\ : \implies \text{First \: year \: investment}( a_{1}) = 500\:rupees\\ \\ : \implies \text{Second \: year \: investment}( a_{2}) = 900\:rupees\\ \\ : \implies \text{Time(n) = 22} \\ \\ : \implies \text{Common \: difference (d)= 400} \\ \\ \red {\underline \bold{To \: Find: }} \\ : \implies \text{Total \: amount \: in \: 22 \: years}( S_{22}) = ?$

• According to given question :

$: \implies S_{n} = \frac{n}{2} (2a + (n - 1) \times d) \\ \\ : \implies S_{22} = \frac{22}{2} (2 \times 500 + (22 - 1) \times 400) \\ \\ : \implies S_{22} =11 \times (1000 + 8400) \\ \\ : \implies S_{22} =11 \times 9400 \\ \\ \green{ : \implies S_{22} = \text{53400 \: rupees}} \\ \\ \green{\therefore \text{Total \: amount \: invested \: in \: 22 \: years = 1;03,400 \: rupees}}$

Answer 2

Answer:

☆ Total Amount = 1,03,400

Step-by-step explanation

□ Using formula of sum of nth terms.

》Sn = n/2[2a+(n-1)d]

• a=500

• d=400

• n=22

=) Sn = 22/2[2 × 500 + (22-1) × 400]

=) Sn = 11 × (1000 + 8400 )

=) Sn = 11 × 9400

=) Sn = 1,03,400 rupees

therefore Total amount invested at the end of 22 years is 1,03,400 rupees