Arithmetic and Geometric progressions?

Question

I get stuck for both AP n GP when the first term isnt given. EG

1. The second term of an AP is 15 and the fifth term is double the first. Find the sum of the first 20 terms of the series.
(a)780 (b)810 (c)720 (d)790 (e)805

2. Find the series in the GP whose 5th term is 80 and the 8th term is 640.
(a)5, 10, 20, 50 (b)2, 4, 6, 8 (c)5, 7, 9, 11 (d)5, 10, 20, 40 (e)5, 20, 40, 80

I figured the answer to the second one is D. I need to know whether this is based on trial and error alone or is there a specific formula to be used here? How do u know what's the difference between the terms or the ratio if only the second term and/or some other random term is given??

thanx!! n what abt the second one?? suppose the options weren't given. How would you solve that?

Answer 1

(1) For an AP, let's call the common difference "x" and the first term "F". You know:

15 = F + x (the second term is 15)
2F = F + 4x (the fifth term is twice the first term)

The second equation reduces to F = 4x, so substitute that into the first:

15 = F + x
15 = 4x + x
3 = x

Now you know the common difference is 3. Since the second term is 15, you know that the first term is 12.

The 20th term is 12+(3*19) = 69, and the sum of the first 20 terms is the average of the first and last terms (12 + 69)/2 times the number of terms (20):

(12 + 69)/2 * 20 =
81 * 10 =
810

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(2) For a GP, the difference between the 5th term and the 8th term is the cube of the common factor. (Let's call x the common factor, and V the fifth term. The sixth term would be x*V, the seventh x^2*V and the eighth x^3*V).

The eighth term divided by the fifth term eliminates V and yields x^3, the cube of the common factor.

640 / 80 = 8, and the cube root of 8 is 2, so the common factor is 2.

If 80 is the fifth term, you can divide by the common factor (which you now know to be 2) going backwards:

80/2 = 40 is the fourth term, and
40/2 = 20 is the third term, and
20/2 = 10 is the second term, and
10/2 = 5 is the first term

Your series is 5, 10, 20, 40, which is choice (d).

Alternatively, you could have reached d by process of elimination, since none of the others is even a geometric progression:

(a) The second term is twice the first, but the fourth is two-and-a-half times the third, so there's no common factor.
(b) The second term is twice the first, but the fourth is one-and-a-third times the third, so there's no common factor.
(c) The second term is 1.4 times the first, and the fourth is one-and-two-ninths tims the third, so there's no common factor.
(e) The second term is four times the first, and the fourth is twice the third, so there's no common factor.

... however that only works because they didn't give better choices. The problem would have been trickier if some of the wrong answers had been things like 10, 20, 40, 80, and 2.5, 5, 10, 20

Answer 2

1. Let a be the first term and the common difference be d.

Tn = a + (n-1)d.

T2 = a+ 1*d = 15 ---------------- 1
T5 = 2*T1 = 2a
But,
T5 = a + 4d;
2a = a + 4d;
a = 4d;

Substituting in 1;
5d = 15;
d = 3.
a = 12.

Sum of first n terms = n*(2a + (n-1)d)/2;
Sum of first 20 terms = 20*(2*12 + (20-1)3)/2;
= 10*(24 + 57);
= 810.
The secret of doing these types of questions is to use an equation connecting all the given values. Don't try to think of all the steps before doing the question.

2.Let a be the first term and the common ratio be r.
The nth term is ar^(n-1).

ar^4 = 80 ----------------- [1]
ar^7 = 640 ----------------- [2]

[2]/[1] =
r^3 = 8;
r = 2.

Substituting in [1],
a * 16 = 80.
a = 5.

The series is
a , ar , ar^2 , ar^3.
5 , 10 , 15 , 20 , .........

You can do better with these questions if you try a step-by-step manner.

Answer 3

I call the difference between the terms a "step" (I use the symbol "s" instead). If I name "a" one of the AP terms, then the first term is a1, the second a2, etc.

Available Data : a2 = 15 ; a5 = 2 x a1
Question: Sumof(a(n)) with n=1 to 20

Solution:
a2 = a1 + s <=> s = a2 - a1 <=> s = 15 - a1 (I name this relation "A")
From a2 to a5 there are 3 steps, so...
a5 - a2 = 3s <=>
2 x a1 - 15 = 3s <=> (Also from "A" we take...)
2 x a1 - 15 = 3(15- a1) <=>
2 x a1 - 15 = 45 - 3 x a1 <=> ( and solving for a1...)
5 x a1 = 60 <=> a1 = 60 / 5 <=> a1 = 12

So, from "A", s = 15 - 12 = 3

The sum of the first 20 terms will be a1 + (a1 + s) + (a1 + 2s) + (a1 + 3s) +...+ (a1 + 19s) = 20 x a1 + 190 x s = 20 x 12 + 190 x 3 = 810

Answer: (b)

Answer 4

let the first term be 15 - d. .... d is the common difference.©
2nd term: 15
3rd term: 15 + d
4th term: 15 + 2d
then the 5th term is 15 + 3d = 30 - 2d.

Thus it turns out that d = 3 and the first term is 12.
Sum: 10/2 * [2*12 + 19*(3)] = 810.


2.
a*r⁴ = 80 .... this is the fifth term. a is the first. r is the common ratio.
a*r⁷ = 640.

a*r⁷ / a*r⁴ = 640/80
r³ = 8.
r = 2.
4th term: 40 ...... 3rd term: 20......... 2nd term: 10.... 1st term: 5