solution to question paper 1 --- IPM STD 5

What is the sum of the number in the series 3, 7, 11, 15, ...., 75

Answer:

I am attaching a small explanation. in case you are interested to know why we are doing whatever we are doing in the way it is done.

We find that each number differs from the other by 4. We will note this difference as “d”

Now say, the middle number of this series is “a”.

So there are as many numbers before the middle number as there are after.

......( a – 4d), (a – 3d), (a – 2d), (a – d), a, (a + d), (a + 2d), (a + 3d), (a + 4d) ........

Sum of all this numbers = S

= ......( a – 4d) + (a – 3d) + (a – 2d) + (a – d) + a +(a + d) + (a + 2d) + (a + 3d) +(a + 4d) ........

We will find that the –d terms will cancel out with +d terms on the other side of “a”

So, S= na, where the “n” = total number of terms.

But we still can’t solve the problem, because we do not know the value of “a” and we do not know “n”.

Let us see if we can find “a”.

We know the first term as say t1 = a – (n/2)d and the last terms as tn = a + (n/2)d

Adding the first and the last term we find t1 + tn = a –(n/2)d  + a + (n/2)d = 2a

So, a = (t1 + tn)/2

Also, tn = t1 + (n-1)d

Or n = ^{(tn – t1)}/_d + 1

Answer: 

The question now becomes simple

 We have,  a = (t1 + tn)/2   = (3 + 75)/ 2 = 39

And, n = ^{(tn – t1)}/_d + 1  = (75 – 3)/ 4 + 1 = ( 72/4) + 1 = 19

So, S = na = 19 x 39 = 741