# In the following Question, Rs. 33,000 is paid off in 12 instalments, such that each installment is Rs. 100 more than the preceding one. Find the amounts of the first and the last instalments.

Last updated date: 15th Mar 2023

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Answer

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Hint: The given question is in A.P., therefore we can consider the value of the number of terms to be 12, common difference to be 100 and then we can use these values to find the sum of n terms using the formula.

Complete step-by-step answer:

${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Let, since we do not know the value of the first instalment, let the first instalment be a.

n = 12

d = 100

${S_{12}} = \dfrac{{12}}{2}\left( {2a + 11d} \right)$

$33000 = 6\left( {2a + 1100} \right)$

$2a + 1100 = 5500$

\[2a = 4400\]

\[a = 2200\]

So, the first instalment is Rs. 2200

Last instalment is 2200+11(100) = 2200+1100=3300

Note: In these questions where there is a common difference we can directly apply arithmetic progression formulas to find the answer.

Complete step-by-step answer:

${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Let, since we do not know the value of the first instalment, let the first instalment be a.

n = 12

d = 100

${S_{12}} = \dfrac{{12}}{2}\left( {2a + 11d} \right)$

$33000 = 6\left( {2a + 1100} \right)$

$2a + 1100 = 5500$

\[2a = 4400\]

\[a = 2200\]

So, the first instalment is Rs. 2200

Last instalment is 2200+11(100) = 2200+1100=3300

Note: In these questions where there is a common difference we can directly apply arithmetic progression formulas to find the answer.

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