- Evaluate
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Find the sum of the A.P. -7-3+1+… from the 7th to the 13th term inclusive.
From the AP it can be seen that a=-7, d=4 and.
The required sum is given by.
Using and, we obtain.
- The first and last terms of an A.P. with 25 terms are 29 and 179. Find the sum of the series and the common difference.
a=29, l=179 and n=25, therefore.
Rearranging the formula to make d the subject, we obtain.
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The sum of the first four terms of an A.P. is twice the 5th term. Show that the common difference is equal to the first term.
From the question we know that, where and. Therefore,.
- Three numbers in an A.P have sum 33 and product 1232. Find the numbers.
Let the numbers be a, b and c.
From the definition of an A.P. we know that a=b-d and c=b+d.
Thus,.
Also
The numbers are 8, 11 and 14.
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In an A.P. the sum of the first 2n terms is equal to the sum of the next n terms. If the first term is 12 and the common difference is 3, find the non-zero value of n.
From the question we know that, a=12 and d=3.
Hence, using
and
Equating the sums
The non-zero value of n is n=7.
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Write down the nth term and the stated term of the following G.P's.
The nth term of a G.P. is
- Find the number of terms in each of the following G.P's and the sum of the series.
The number of terms is giving by solving the last term for n and the sum of the series is .
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From inspection we see that a=1/4, l=64, r=2 and.
The number of terms is giving by solving for n.
Thus,.
Then,
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a=1000, l=0.32 and r=1/5. The nth term is.
and so n=6.
The sum is .
- The first term of a G.P. with positive terms is 80. If the sum of the first three terms is 185, find the common ratio.
a=80 and.
As the question states that the series has positive terms, only the positive value of r is valid. So r=3/4.
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Find two distinct numbers p and q such that p, q, 10 are in arithmetic progression and q, p, 10 are in geometric progression.
For the A.P.
.
Similarly, for the G.P.
Substituting into, we obtain
We ignore the value of p=10 and q=10, which are not valid as p and q must be distinct values.
Putting p=-5 into the equation gives q=2.5.
Thus, p=-5 and q=2.5.
- In a G.P. the second term exceeds the first by 20 and the fourth term exceeds the second by 15. Find the two possible values of the first term
We are given
As the series is a G.P. the terms can be written as
Making r the subject in the first equation and substituting into the second
Solving for a gives a=-40 or a=-8.
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The sum of (n+12) terms of the G.P. 2+4+8+… is twice the sum of the n terms of the G.P. 3+12+48+… Calculate the value of n.
In order to be able to tell the different series apart, let the sum to n of the first series be and the sum of the second series be.
In the first G.P. a=2 and r=2, so.
In the second G.P. a=3 and r=4, so.
Equating the sums
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Find the sum of the first n terms of the G.P. 1/12+1/4+3/4+… How many terms are required for the sum to be greater than 100?
a=1/12, r=3 and.
For we have
Therefore, we need at least 8 terms for the sum to be greater than 100.
- Determine whether the series below converge. Find the sum to infinity of those that are convergent.
An infinite series converges if the common ration r satisfies. The sum to infinity is.
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This is a G.P. with a=4 and. It is convergent.
- This is an Arithmetic Progression and so does not converge for infinite terms.
- This is a G.P. with a=20 and r=-0.5. It is convergent.
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This is a G.P. with and . It is convergent.
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This is a G.P. with a=3 and . It is convergent.
- A geometric series with common ratio 0.8 converges to the sum 250. Find the fourth term of the series.
The sum of a geometric series is
r=0.8 and S=250, thus
The fourth term is given by.
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The sum of the first n terms of a geometric series is. Find the first term of the series, its common ratio and its sum to infinity.
If two adjacent terms of a G.P. are known then
The first term is.
The second term is.
Thus,.
The sum to infinity is.
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Find the sum to n terms of the geometric series 108+60+33 1/3 +… If k is the least number which exceeds this sum for all values of n, find k. Find also the least value of n for which the sum exceeds 99% of k.
a=108 and.
The sum of n terms is.
The value of k is the smallest number bigger than for all values of n. Thus, .
The least value of n for which the sum exceeds 99% of k is given by the solution of
Taking logs and noticing that (this means that when we divide, the greater than sign is reversed)
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Show that
