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Sequences and Series Ncert Solutions Maths class 11th Maths Class 11th Sequence and Series

104. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

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Answered by

Payal Gupta | Contributor-Level 10

6 months ago

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 \times 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal\timesrate \timestime}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 \times 5 \times 2}{100}) {\frac{Principal\times rate\timestime}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 \times 5 \times 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) × 500

=? 10500 + 13 × 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1)

...more

104. Given,

Principal amount =? 10000

Amount at end of 1 year =? $(10000 + \frac{10000 + 5 \times 1}{100})$

=? (10000 + 500)

=? 10500 {Amount paid = principal + S.I. in a year}

$S . I = \frac{Principal\timesrate \timestime}{100}$

Amount at end of 2nd year

=? $(10000 + \frac{10000 \times 5 \times 2}{100}) {\frac{Principal\times rate\timestime}{100}}$

=? (10000 + 1000)10500 + 500

=? 11000

Amount at end of 3rd year

=? $(10000 + \frac{10000 \times 5 \times 3}{100})$

=? (1000 + 1500)

=? 11500

So, amount at end of 1^st, 2^nd, 3^rd ………, n^th year forms as A.P.

i.e.? 10500? 11000? 115000, ………. with

a = 10500

d = 11000 - 10500 = 500

Now, Amount in 15th year = Amount at end of 14th year

=? 10500 + ( 14 - 1) × 500

=? 10500 + 13 × 500

=? 10500 + 6500

=? 17000

Similarly amount after 20th year, =? 10500+ (20 - 1) × 500

=? 10500 + 19 × 500

=? 10500 + 9500

=? 20,000

less

104. Given, Principal amount =? 10000Amount at end of 1 year =?  <math><mrow><mrow><mo> (</mo><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>+</mo><mfrac><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>+</mo><mn>5</mn><mo>×</mo><mn>1</mn></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math>=? (10000 + 500)=? 10500 {Amount paid = principal + S.I. in a year}<math><mrow><mi>S</mi><mo>.</mo><mi>I</mi><mo>=</mo><mfrac><mrow><mo>Principal×rate</mo><mo>×time</mo></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac></mrow></math>Amount at end of 2nd year=?  <math><mrow><mrow><mo> (</mo><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>+</mo><mfrac><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>×</mo><mn>5</mn><mo>×</mo><mn>2</mn></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac></mrow><mo>)</mo></mrow><mrow><mo> {</mo><mrow><mfrac><mrow><mo>Principal×</mo><mo>rate×time</mo></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac></mrow><mo>}</mo></mrow></mrow></math>=? (10000 + 1000)10500 + 500=? 11000Amount at end of 3rd year=?  <math><mrow><mrow><mo> (</mo><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>+</mo><mfrac><mrow><mn>1</mn><mn>0</mn><mn>0</mn><mn>0</mn><mn>0</mn><mo>×</mo><mn>5</mn><mo>×</mo><mn>3</mn></mrow><mrow><mn>1</mn><mn>0</mn><mn>0</mn></mrow></mfrac></mrow><mo>)</mo></mrow></mrow></math>=? (1000 + 1500)=? 11500So, amount at end of 1st, 2nd, 3rd ………, nth year forms as A.P.i.e.? 10500? 11000? 115000, ………. witha = 10500d = 11000 - 10500 = 500Now, Amount in 15th year = Amount at end of 14th year=? 10500 + ( 14 - 1) × 500=? 10500 + 13 × 500=? 10500 + 6500=? 17000Similarly amount after 20th year, =? 10500+ (20 - 1) × 500=? 10500 + 19 × 500=? 10500 + 9500=? 20,000

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104. A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

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