A man arranges to pay off a debt of $3600 by 40 annual installments wh

An arithmetic sequence has the property median = mean. The first 40 terms has a median of \(\frac{a(1) + a(40)}{2}\), the first 30 terms has a median of \(\frac{a(1) + a(30)}{2}\). We can use these to replace the mean, hence build the sum using the median * number of terms.

First 40 installments: \(\frac{a(1) + a(40)}{2} * 40 = 3600\), \(a(1) + a(40) = 180\).
First 30 installments: \(\frac{a(1) + a(30)}{2} * 30 = 2400\), \(a(1) + a(30) = 160\).

Then we can eliminate a(1) to get: \(a(40) - a(30) = 20\). \(a(40) - a(30)\) is also the difference of 10 terms, so the differerence between consecutive terms is 20/10 = 2.

Finaly \(a(1) + a(30) = a(1) + a(1) + 29*2 = 160\) and \(a(1) +29 = 80\), \(a(1) = 51\). Add 7 differences which is 14 to get 65.

Ans: C

So amt. paid in the 30 first installments is 2400
Hence \(\frac{30}{2}\)\((x+x+29(d)) = 2400\)
\(2x+29(d)= 160\) (i)

Also amt. paid in the last 10 installments is 1200 this can be expressed as
The sum from \(31^{th}\) to \(40^{th}\) term is 1200
so\( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)
after simplification we get :
\(2x+69d=240\) (ii)


Solving (i) an (ii)
We get d= 2 and x= 51

The value of the eight term is \(x + 7d\) where \(x\) is the first term and d is the common difference of the A.P.
Putting the values of x and d we get : 51+7(2)= 65

Ans-C

Hope it's clear.

Hi stne, could you please explain how you obtained the LHS of the formulas \(\frac{30}{2}\)\((x+x+29(d)) = 2400\) & \( \frac{10}{2}( x+30(d) + x+ 39(d))=1200\)? Thanks!

Hi lucasd14,

You need to know TWO basic properties of terms that are in A.P:

(1) The sum of n terms in an A.P. is given by \(\frac{n}{2}\){ 1st term + last term}...(i)

Also any given term in an AP can be found if we know the first term and common diff.

(2) nth term= a + (n-1)d ...(ii)

Lets say we need the \(5\)th term
Lets say \(a \) is the first term
lets say \(d \) is the diff. between consecutive terms

Hence the \(5\)th term:
\(= a + (5-1)d\)
\(= a+4d \)

Coming back to the question :

\(\frac{1}{3}*3600\) was unpaid so \(1200 \) was unpaid .

Hence he paid remaining \(2400\) in \(30\) installements which were in A.P.

Now that you know the sum of n terms that are in A.P is given by \( \frac{n}{2}\){ 1st term + last term} you need to plug in the values

\(n = 30\)
let \(x= 1\)st term
let the diff. between two terms be \(d\)
Last term \(( 30\)th term ) \(= x+ (30-1)d\)... using (ii)
Now we know the sum of the \(30\) installements \(= 2400\)

Hence \( \frac{30}{2}\)\(({ x+ x+ (30-1)d }) = 2400 \)... using (i)

\( \frac{30}{2}\) \(({ x+ x+ (29)d }) =2400\)

Now in the same way you should be able to get the second part of your query:

In this part we are finding the sum of \(10\) terms (that are in A.P.)and the sum equals \(1200\)
\(1\)st term is the \(30\)th term
Last term is the \(40\)th term
\(1\)st term= \(30\)th term = \(x+ 29\)d ...( using ii)
Last term =\(40\)th term\(= x + 39\)d ...(using ii)

Hence finally \(\frac{10}{2}(x+ 29d + x + 39d) = 1200\)...using (i)

Should be clear, if anything is still unclear, feel free to ask. All the best.

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