Sum of a number of terms in a.p., Mathematics

Assignment Help:

We know that the terms in an A.P. are given by

a, a + d, a + 2d, a + 3d, ........ a + (n - 2)d, a + (n -  1)d

The sum of all these terms which is denoted by "S" is given by

  S = n/2 {2a + (n - 1)d}

This is obtained as follows. We know that

S       =       (a) + (a + d) + (a + 2d) + (a + 3d) + ..... +

                   {(a + (n - 2)d)} + {(a + (n - 1)d)}

Now we reverse the order and write it as shown below.

S       =       (a + (n -1) d) + (a + (n - 2) d) + ......... +

                   (3d + a) +  (2d + a) + (d + a) + a

On adding the respective terms we get

2S     =       {a + a + (n - 1)d} + {a + d + a + (n - 2)d}

                   + ......... + {a + (n - 2)d  + a + d} +

                   {a + (n - 1)d + a} 

That is, we have:

2S     =       {2a + (n - 1)d} + {2a + d + (n - 2)d} +

                    ............... + {2a + d + (n - 2)d} +

                   {2a + (n - 1)d}

Further simplifying we obtain

2s     =       {2a + (n - 1)d} + {2a + d + nd - 2d} +.............. +

                   {2a + d + nd - 2d} + {2a + (n - 1)d}

On simplification we obtain

2s     =       {2a + (n - 1)d} + {2a + nd - d} + ......... +

                   {2a + nd - d} + {2a + (n - 1)d}

2s     =       {2a + (n - 1)d} + {2a + (n - 1)d} + ....... +

                   {2a + (n - 1)d} + {2a + (n - 1)d}

Since 2 + 2 + 2 + 2 = 2(1 + 1 + 1 + 1) = 2 x 4,

2a + (n - 1)d  multiplied n times will be  n.{2a + (n - 1)d}. Therefore,

         2s      =       n.{2a + (n - 1)d}

                            or

s

= n/2 {2a + (n - 1)d}       ............. (a)

Since l = a + (n - 1)d,  equation (a) is also written as

s

= n/2   {a + a + (n - 1)d}  or       

= n/2 {a + l}

Now we will find the sum of 20 terms when a = 5 and d = 2. Substituting these values in the formula, we obtain

s

= 20 /2 {2(5) + (20 - 1)2}
  = 480  

This problem can also be solved by finding the last term which in this case happens to be T20 and it is given by T20 = 5 + (20 - 1)2 = 43. Therefore,

s = n /2 {a + l}
  = 20 /2 {5 + 43} = 480.

We observe that both these methods are essentially the same. With this background let us look at few more examples.

Example 

For the series given below, find the 23rd and the 27th terms.

                   38, 36, 34, .............

We are given the first term that is a = 38. The common difference d is given by 36 - 38 = -2.  The 23rd term is given by

         T23    =       a + 22d

                   =       38 + 22(-2)  

                   =       38 - 44  = - 6  

Similarly the 27th term is given by

         T27    =       a + 26d

                   =       38 + 26(-2)

                   =       38 - 52


Related Discussions:- Sum of a number of terms in a.p.

Error analysis: describle and correct the error in plotting, to plot (5,-4)...

to plot (5,-4), start at (0,0) and move 5 units left and 4 units down

Conclusion of egroff''s theorem and lusin''s theorem, (1) Show that the con...

(1) Show that the conclusion of Egroff's theorem can fail if the measure of the domain E is not finite. (2) Extend the Lusin's Theorem to the case when the measure of the domain E

Determine the numbers of sides in regular polygon, If each interior angle o...

If each interior angle of a regular polygon has a calculated as of 144 degrees, Determine the numbers of sides does it have? a. 8 b. 9 c. 10 d. 11   c. The measur

I need help in math badly, I just have a hard time in math in every other c...

I just have a hard time in math in every other class I have an A or B but in math I have a C+ I at least want a B- or B+ or A- or even an A+

5th grade, 6 and 3/8 minus 1 and 3/4

6 and 3/8 minus 1 and 3/4

Estimation of difference among two means-illustration, A comparison of the ...

A comparison of the wearing out quality of two types of tyres was obtained by road testing. Samples of 100 tyres were collected. The miles traveled until wear out were recorded and

Prove that xa+ar=xb+br of circle, In figure, XP and XQ are tangents from X ...

In figure, XP and XQ are tangents from X to the circle with centre O. R is a point on the circle. Prove that XA+AR=XB+BR Ans:    Since the length of tangents from externa

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd