Aptitude: Sequence and Series

Formulas:

Nth term of the AP = T_n = a + (n – 1)d
Sum of the first n terms of an AP is S_n = n/2(2a + (n – 1)d) or n/2(a + l),where l is the last term.
Arithmetic mean of two numbers a and b can be found by M = a + b / 2
In GP the common ratio can be found out by r = second term/ first term.
Nth term of GP can be given by T_n = ar^n-1 = l.
Sum of n terms of a GP is S_n = a(rⁿ – 1)/r – 1 if r > 1 and S_n = a(1 – rⁿ)/ 1- r, if r < 1.
Sum of infinite terms is S_n = a/1-r.
Geometric mean is given by G=√ab, a>0 b>0.
The terms are in Harmonic progression if its reciprocal are in AP.
Nth term in HP is t_n = 1/((1/a1)(n-1)(1/a2 – 1/a1)).
HM is found by HM = 2ab/a+b.
A >= G >= H
G² = AH.

Practice Problems:

Problem: If a, b, c are in AP p, q, r are in HP and ap, bq, cr are in GP, then p/r + r/p is equal to

Solution:

Since a, b, c are in AP b = (a + c) / 2

p, q, r are in HP q= 2pr/p+r

And ap, bq, cr are in GP , b² q² = apcr

a+c/2)(4(pr)²/(p+r)²) = apcr

(a+c)² /ac = (p + r)²/pr => p² /pr + r² / pr + 2 = a² /ac + c²/ac + 2

p/r + r/p = a/c + c/a

Problem: If 3 + 5 + 7 + … + n/5 + 8 + 11 +…+10 terms = 7, then the value of n is

Solution:

n/2[2×3 + (n – 1)2]/ (10/2) [2×5 + (10-1)3] = 7

n(n+2)/5×37 = 7

n² + 2n – 1295 = 0 => (n+37)(n-35) = 0 => n = 35.

So the value of n is 35.

Problem: If sum of n terms of an AP is 3n2 + 5n and Tm = 164 then m is equal to

Solution:

T_m= S_m – S_m-1

164 = 3(2m – 1)+ 5.1

6m = 162 => m = 27

Problem: In a GP, if the (m+n)th term be p and (m-n)th term be q, then its mth term is

Solution:

T_m+n = ar^m+n-1 = p, T_m-n= ar^m-n-1 = q

On mul both we get a²r^2m-2 = pq

T_m = ar^m-1 = √pq

Problem: The sum of the first n terms of the series 1/2 + 3/4 + 7/8 + 15/16 + … is

Solution:

½ + ¾ + 7/8 + 15/16 +…+ n terms

= (1 – 1/2)+(1 – 1/4)+(1- 1/8)+….+(1-1/2ⁿ)

= n – (1/2 + 1/4 + 1/8 +…+ 1/2ⁿ)

= n – 1/2(1 – (1/2)ⁿ/1-1/2) = n + 2^-n – 1

So the final answer is (n + 2^-n – 1)

Problem: An AP consists of n (odd terms) and its middle term is m then the sum of the AP is

Solution:

Middle term = T_n+1/2

a+(n+1 / 2 -1 )d = m

2a + (n-1)d = 2m

S_n = n/2[2a + (n-1)d] = nm

Problem: The sum of 1 + 2/5 + 3/52 + 4/53 +…. Infinite terms is

Solution:

here a direct formula is used if AP and GP occurs simultaneously.

Here r=1/5, d=1

S = a/1-r + dr/(1-r)²

=1/(1-(1/5)) + 1×1/5 / (1-1/5)²

=5/4 + 5/16

= 25/16

Problem: The sum of n terms of an AP is an(n-1). The sum of the squares of these terms is equal to

Solution:

S_n = an(n-1), then S_n-1 = a(n-1)(n-2)

T_n= S_n – S_n-1 = 2a(n-1)

T_n² = 4a²(n-1)²

Sum = ∑T_n² = 4a²(n-1)n(2n-1)/6 2a²n(n-1)(2n-1)/3

Problem: If S be the sum to infinity of a GP, whose first term is a, then the sum of first n terms is

Solution:

let r be a common ratio of GP, then

S=a/1-r , r = 1-a/S

S_n = a(1-rⁿ)/1-r

= S[1-(1-a/S)ⁿ]

Problem: If the non-zero numbers a, b, c are in AP and tan-1a, tan-1 b, tan-1 c are also in AP, then

(a). a = b = c	(b). b²= 2ac
(c). a² = bc	(d). c² = ab

Solution:

since 2b= a + c and 2tan^-1b = tan^-1 a + tan^-1 c

2b/1-b²= a + c/1 – ac => b²= ac

4b² = 4ac => (a + c)²– 4ac = 0

(a – c)² = 0

a = b = c

Problem: 1/(b-a) + 1/(b-c) = 1/a + 1/c, then a, b, c are in AP, GP or HP

Solution:

1/(b – a) – 1/c = 1/a – 1/(b-c)

(c – b + a) / c(b – a) = (b – c – a) / a(b – c)

1/c(b – a) = -1/a(b – c) => ba – ca = -cb + ac

ab + bc = 2ac => b = 2ac/(a + c)

so the terms are in HP.

Problem: The value of x + y + z is 15, if a, x, y, z, b are in AP while the value of 1/x + 1/y + 1/z is 5/3, if a, x, y, z, b are in HP, then, a and b are

Solution:

As a, x, y, z, b are in AP then,

X + y + z = 3((a + b)/2)

15 = 3 (a + b/2)

a + b = 10

Also, a, x, y, z, b are in HP

1/a, 1/x, 1/y, 1/z, 1/b are in AP.

1/x + 1/y + 1/z = 3(a + b)/2ab => 5/3 = 3 x 10/2ab

ab = 9 on solving these equations we get a= 1, b=9 or b=1, a=9.