PROBLEMS ON ARITHMETIC SERIES

Problem 1 :

The sum of first n terms of a certain series is given as

3n²-2n

Show that the series is an arithmetic series.

Solution :

S_n  =  3n²-2n

a+a+d  =  8

2a+d  =  8

2(1)+d  =  8

d  =  6

let us find the sum of (n-1) terms using Sn

S_(n-1)  =  3(n-1)²-2(n-1)

t_n  =  S_n-S_(n-1) 

=  3n²-2n-[3(n-1)²-2(n-1)]

t_n = 3n²-2n-[3(n²+1-2n)-2(n-1)]

=  3n²-2n-[3n²+3-6n-2n+2]

=  3n²-2n-3n²-3+6n+2n-2

=  3n²-2n-3n²-3+6n+2n-2

=  6n-5

=  6n-6  

=  6(n-1)+1

=  1+(n-1)6

This exactly matches with the form a + (n-1) d.

So, the first term is 1 and common difference is 6.

Problem 2 :

Show that the sum of an arithmetic series whose first term is a, second term is b and the last term is c is equal to

[(a+c) (b+c-2a)]/2(b-a)

Solution :

From this series we know that the first term is a ,the second term is b and the last term is c

a = a     d = b-a  and L= c

Now we have to to find the value of n

So, n  =  [(l-a)/d] + 1

=  [(c-a)/(b-a)] + 1

=  (c-a+b-a)/(b-a)

=  (c+b-2a)/(b-a)

Now we have to use the formula for Sn to find sum of n terms

S_n =(n/2)(a+l)

=  {[(c + b - 2a)/(b-a)]/2}[a+c]

=  [(c + b - 2a)(a+c)]/2(b-a)

Problem 3 :

If there are (2n+1) terms in an arithmetic series, then prove that the ratio of the sum of odd terms to the sum of even terms is (n+1) : n

Solution :

Let T and S are sum of odd and even numbers respectively.

T  =  t₁ + t₃ + t₅ + ........... + t_(2n+1)

there are (n+1) terms.

To find its sum we have to use S_n formula but here S_n is T_n

T_n  =  (n/2) [a+l]

=  [(n+1)/2] [t₁+t_(2n+1)]

=  [(n+1)/2] [a+a+(2n+1-1)d]

=  [(n+1)/2] [a+a+(2n+1-1)d]

=  [(n+1)/2] [2a+(2n)d]

=  (n+1) (a+nd)  

S = t₂ + t₄ + t₆ + ........... + t_2n

S_n  =  (n/2) (a+l)

= (n/2)[t2 + t 2n]

  =  (n/2)[(a+d) + a+(2n-1)d]

=  (n/2)[a+d + a + 2nd-d]

=  (n/2)[2a+2nd]

=  n [a + n d]

=  Sum of odd terms : Sum of even terms

=  [(n+1)] [ a + n d ] :  n[a + n d]

=  [(n+1)] [ a + n d ]/n[a + n d]

=  [(n+1)]/n

=  (n+1):n

Problem 4 :

The ratio of the sums of first m and first n terms of an arithmetic series is m²:n² show that the ratio of the m th and nth terms is (2m-1) :(2n-1)

Solution :

Sum of m terms : Sum of n terms  =  m² : n²

(m/2)[2a+(m-1)d]/(n/2)[2a+(n-1)d]  =  m² / n²

(m/2) ⋅ (2/n) [2a+(m-1)d]/[2a+(n-1)d]  =  m² / n²

(m/n) [2a+(m-1)d]/[2a+(n-1)d]  =  m²/n²

[2a+(m-1)d]/[2a+(n-1)d]  =  m/n

n[2a+(m-1)d]  =  m[2a+(n-1)d]

2na+ n(m-1)d  =  2ma + m(n-1) d

2na+ nmd-nd  =  2ma+mnd-md

2na-2ma  =  nd-md

2a(n-m)  =  d(n-m)

2a  =  d

Now we have to verify that the ratio of the m^th and n^th terms is (2m-1) :(2n-1)

t_m : t_n = [a+(m-1)d]/[a+(n-1)d]

=  [a+(m-1)(2a)]/[a+(n-1)(2a)]

=  [a + 2ma-2a]/[a + 2na-2a]

=  [-a + 2ma]/[-a + 2na]

=  a [2m - 1]/a[2n - 1]

=  [2m - 1]/[2n - 1]

=  [2m - 1]:[2n - 1]

Problem 5 :

A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end  and the construction should stop at 25th row. How many bricks does he need to buy?

Solution :

If we write the number of bricks in each row as sequence we will get

97, 93, 89,............

Now we nee to find number of bricks needed to buy. For that we have to make it as series and we have to find for 25 terms of the series.

S_n  =  (n/2)[2a+(n-1)d]

a = 97, d = 93-97 ==> -4, n = 25

S₂₅  =  (25/2)[2(97)+(25-1)(-4)]

=  (25/2) [194 + (24) (-4)]

=  (25/2) [194 - 96]

=  (25/2) (98)

=  25(49)

=  1225 bricks

So, the gardener needs 1225 bricks.

Problem 7 :

The sum of the series -8, -6, -4, ........... n terms is 52. The number of terms n is 

a)  11     b)  12   c)  13    d)  10

Solution :

a = -8, d = -6 - (-8)

d = -6 + 8

d = 2

S_n = (n/2)[2a+(n-1)d]

(n/2)[2a+(n-1)d] = 52

Applying the values of a and d, we get

(n/2)[2(-8)+(n-1)(2)] = 52

(n/2) [-16 + 2n - 2] = 52

(n/2) [-18 + 2n] = 52

(n/2) x 2 [-9 + n] = 52

n(-9 + n) = 52

-9n + n² = 52

n² - 9n - 52 = 0

(n - 13) (n + 4) = 0

n = 13 and n = -4

So, the required number of terms is 13.

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