## 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(r^{n}– 1)/r – 1 if r > 1 and S_{n}= a(1 – r^{n})/ 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
- G
^{2}= 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 , b^{2} q^{2} = apcr

a+c/2)(4(pr)^{2}/(p+r)^{2}) = apcr

(a+c)^{2} /ac = (p + r)^{2}/pr => p^{2} /pr + r^{2} / pr + 2 = a^{2} /ac + c^{2}/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

n^{2} + 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 a^{2}r^{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)+….+(1-1/2^{n})

= n – (1/2 + 1/4 + 1/8 +…+ 1/2^{n})

= 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_{n}^{2} = 4a^{2}(n-1)^{2}

Sum = ∑T_{n}^{2} = 4a^{2}(n-1)n(2n-1)/6 2a^{2}n(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-r^{n})/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). b^{2 }= 2ac |

(c). a^{2} = bc | (d). c^{2} = ab |

**Solution:**

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

2b/1-b^{2}= a + c/1 – ac => b^{2}= ac

4b^{2} = 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.