Question

Find the sum :
(i)2+4+6...+200
(ii)3+11+19+...+803
(iii)(-5)+)-8)+(-11)+...+(-230)
(iv)1+3+5+7+...+199
(v)7+701/2+14+...+84
(vi)34+32+30+...+10
(vii)25+28+31+...+100
(viii)18+151/2+13+...+(-49 1/2)

Answer 1

Answer:

Now We are asked to find sum of some number of terms in Arithmetic progression.

We know that, Sum of n terms = n/2 ( a + l)

where a is the first term, l is the last term.

(i) 7 + 10 1/2 + 14 + ... + 84

Here a = 7

l = 84

d = 3.5

We know that, nth term of this A. P is l

84 = a + ( n - 1 ) d

84 = 7 + ( n - 1 ) 3.5

84 = 7 - 3.5 + 3.5n

84 - 3.5 = 3.5n

80.5/3.5 = n

n = 23 .

So sum of first 23 terms = 7 + 10 1/2 + 14 ... + 84 = 23/2 ( 7 + 84 )

= 11.5 * 91

= 1046.5

(ii) 34 + 32 + 30 + . . . + 10

First term ( a) = 34

Last term ( l) = 10

Common difference ( d) = 32 - 34 = -2

We know that, the nth term of the A. P is l

10 = 34 + ( n - 1 )d

10 = 34 + (n-1)*-2

10 - 34 = -2n + 2

-24 - 2 = -2n

-26 = -2n .

n = 13

Sum of first thirteen terms = 34 + 32 + 30+ ... + 10 = 13/2 ( 34 + 10 ) = 13 * 22 = 286 .

(iii) –5 + (–8) + (–11) + . . . + (–230)

First term( a) = -5

Last term ( l) = -230

Common difference = -8 - ( -5 ) = -3

Let the nth term of this A. P be l

-230 = -5 + ( n - 1 ) d

-230 + 5 = ( n - 1 )* -3

-225 = -3n + 3

-225- 3 = -3n

-228 = -3n

n = -228 / -3 = 76

Sum of 76 terms = –5 + (–8) + (–11) + . . . + (–230) = 76/2 ( -5 -230 ) = 38* -235 = -8930

Conclusion :

(i) 7 + 10 1/2 + 14 .... + 84 = 1046.5

(ii) 34 + 32 + 30 + . . . + 10 = 286

(iii) –5 + (–8) + (–11) + . . . + (–230) = -8930