2n plus 3 is an arithmetic sequence how many terms should be taken starting from the first term so that the ratio of the sum of the first third of these terms to the rest of the terms is 7 to 44?

Answer 1

The sum of an arithmetic series of n terms is:

S(n) = n(2a1 + (n - 1)d) / 2

where a1 is the first term and d is the difference between terms.

The key to solving this is to realise that two sums are required, S1 and S2 such that:

S1 = S(r)

S2 = S(3r) - S(r).

But that this second sum can also be stated as:

S2 = sum of 2r terms starting at term ar+1

So that the problem then becomes:

Find two sums S1 and S2 of the sequence an = 2n + 3 such that the first (S1) has r terms starting at a1 and the second (S2) has 2r terms starting at ar+1, with the ratio of S1 : S2 = 7 : 44.

This gives:

S1 = r (2a1 + (r - 1)d) / 2

= r(2(2x1 + 3) + (r - 1)2) / 2

= r(2r + 8) / 2

= r(r + 4)

S2 = 2r(2ar+1 + (2r - 1)d)

= 2r(2(2(r + 1) + 3) + (2r - 1)2) / 2

= 2r(2(2r + 5) + (2r - 1)2) / 2

= 2r(2r + 5 +2r - 1)

= 2r(4r + 4)

= 8r(r + 1)

As S1 : S2 = 7 : 44,

44 S1 = 7 S2

=> 44 r(r + 4) = 7 x 8r(r + 1))0

=> 11(r + 4) = 14(r + 1)

=> 11r + 44 = 14r + 14

=> 3r = 30

=> r = 10

As r is the number of terms in S1 it is a third of the number of terms required, so 3 x 10 = 30 terms are needed.

To check:

a1 = 2 x 1 + 3 = 5

S(30) = 30(2 x 5 +(30-1) x 2) / 2 = 30 x 34 = 1020

S(10) = 10(2 x 5 + (10-1) x 2) / 2 = 10 x 14 = 140

Sum of 1st third S1 = S(10) = 140, sum of rest S2 = S(30) - S(10) = 1020 - 140 = 880

Ratio of S1 : S2 = 140 : 880

= 14 : 88

= 7 : 44

Note that the second sum S2 is also 20 terms starting at a11 = 2 x 11 + 3 = 25:

S2 = 20(2 x 25 + 19 x 2)/2

= 20 x 44

= 880.