Progressions (AP, GP & HP)

Finite Series :- Series with finite number of terms.

Infinite Series :- Series with infinite number of terms.

Arithmetic Progression (A.P.):- Series with difference of consecutive terms is constant. 

e.g.  7, 11, 15, 19 .... 

First term :- a₁   = 7

Common Difference = d = 3

then nth term of series (a_n) = a₁ +(n-1)d 

&

sum of first 'n' terms of A.P. (Sn) =  n(2a₁ +(n-1)d ) / 2   or  n(a₁ +a_n ) / 2 

Q1. Find the 8th term and sum of first 10 terms of the following series 

5,8,11,14....

(1)23, 184 (2)26, 185 (3)26, 155 (4) 29, 132 (5) 29, 184

Solution:- 

Given series is a A.P. with first term = 5 & common difference = 3.

so 8th term of series = 5 + (8-1)*3 = 5 + 21 = 26

sum of first 10 terms = 10(2*5 + 7*3) / 2 = 155. Hence answer option 3.

Important Notes:-
I. If three numbers are in an A.P. consider: a-d, a, a+d
II. If four numbers are in an A.P. consider: a-3d, a-d, a+d, a+3d
III.Middle term of the an A.P. : Arithmetic mean of the series.

Q2. if 15th term of an A.P. series is 30. Find sum of first 29 terms of same series.
(1) 435 (2) 450 (3) 870 (4) 680 (5) 652

Solution:-
For first 29 terms, 15th  term will be middle term i.e. average of first 29 terms. 
so sum of first 29 terms = 30 * 29 = 870. Hence answer option 3.


Q3. A pentagon has its sides in A.P. If perimeter of pentagon is 80 then find sum of largest & smallest side of pentagon ?
(1) 16 (2) 32 (3) 64 (4) 48 (5) Data insufficient

Solution:-
let five sides  be a-2d, a-d, a, a+d, a+2d
sum of five sides = 5a = 80 
middle side = a = 16
sum of largest & smallest side = a-2d+a+2d = 2a = 32. Hence answer option 2.


Q4. Find sum of all numbers which are divisible by 11 and lying between 120 & 442 on number line.
(1) 8415 (2) 8134 (3) 4208 (4) 2104 (5) None of these

Solution:- 
First number in this range = a = 121
last number = 440
total such numbers = (440-121)/11  + 1 = 30
Sum of numbers of series 121, 132, 143, ...... 440 
= n(a₁ +a_n ) / 2
= 30 (121+440) / 2 = 8415. Hence option 1.


Q5. How many terms in following series must be taken such that their sum is 44.
-11,-8,-5, ......
(1) 8 (2) 9 (3) 10 (4) 11 (5) 12

Solution:- 
(Sn) =  n(2a₁ +(n-1)d 
44 = n(-22 + (n-1) 3)
then n = 11. Hence answer option 4.
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Geometric Progression (G.P.):- Ratio of consecutive terms in constant. 
e.g. 3, 15, 75, 375 ...

First term :- a₁   =3
common ratio = r = 15/3 = 75/15 = 5
then
'n^th' term of a G.P. = ar^n-1

sum of first 'n' terms of G.P. series (Sn) = a (rⁿ- 1)/(r-1)

Q6. For given G.P. 3, 15, 75, ...
Find  4th term & sum of first 4 terms. 


Solution:-
a = 3   & r = 5
4th term = ar^n-1   =  3*5^4-1  = 3*125 =375
sum of first four terms = 3 (625-1) / 4 = 468

Important Notes:-
I. If three numbers are in G.P. consider a/r, a, ar
II. If four numbers are in G.P. consider a/r³ ,a/r , ar , ar³
III. The middle term of a G.P. is geometric mean of the series .


Q7. if three numbers x, y & z are in G.P. in given order. If x = 3/5 & z = 4/15 then find y.
(1) 2/3 (2) 2/5 (3) 3/2 (4) 3/5 (5)5/3

Solution:-
let three numbers be  a/r, a, ar

then z / x = r² = 4 / 9 then r = 2/3
then y = a = x * r = (3/5) * (2/3) = 2/5. Hence answer option 2.


Q8.if x, y & z are in G.P. then log x, log y & log z will be in 


(1) A.P. (2) G.P. (3) H.P. (4) A.G.P. (5) No standard series 


Solution :- 
As we know log (a.b) =  log a + log b
let three terms in G.P. be a/r, a, ar
then log x , log y , log z be log (a/r), log a,log (ar)
= log a - log r , log a , log a + log r.
above series is in A.P. hence option 1.

Q9. if ratio of first 8 terms of G.P. to first four terms of series is 82:1. The find the common difference.

(1) 81 (2) 3 (3) 9 (4) 1/3 (5) 1/9

Solution:-
a (r⁸- 1)/(r-1) ÷ a (r⁴- 1)/(r-1)
=  (r⁸- 1) ÷ (r⁴- 1)
= (r⁴ + 1)(r⁴- 1) / (r⁴- 1)
= (r⁴+ 1) = 82
r⁴ = 81
r = +3 or -3. Hence answer option 2.
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Harmonics Progression (H.P.): Reciprocal terms are in A.P.
e.g. 1/7, 1/11, 1/15, 1/19 ....
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Arithmetic-Geometric Progression (A.G.P.) : (CAT 2004)
--- will put it soon.
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Infinitely diminishing series:
Three mandate requirements are
I. Series must be a G.P.
II. Common ratio between -1 & 1.   -1 < r < 1 & r ≠ 0.
III. Infinite terms.

Sum of all terms of Infinite series = a / (1-r)

Q10. Find the sum of all terms of following series. 
81, 27, 9, 3, 1, 1/3, 1/9, 1/27, ....... Infinite terms.

(1) 120 (2) 121.5 (3) 122.5 (4) 242.5 (5) 243 

Solution:-
Given series is infinitely diminishing series with a = 81 & r = 1/3.

Sum of series = a / (1-r) = 81 / (1-1/3) = 81 * 3 /2 = 243 / 2 = 121.5.
Hence answer option 2.
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