Question

Find the number of terms of the series 21,18,15,12....which must be taken to give a sum zero

Answer 1

a = 21
d = a2-a1
   = 18-21
   = -3
an = 0
                    an = a + (n - 1) d
                     0  = 21 + (n - 1) -3
                     0  = 21 -3n +3
                     0  = 24 - 3n 
                    3n = 24
                      n = 8
therefore ,
                  8 no.s of the series should be taken to give a sum to 0 .   

Answer 2

Answer:

The number of terms = 15

Step-by-step explanation:

Required to find,

The number of terms such that the sum to n-terms of the series 21,18,15,12.... is 0

Recall the formula,

The sum to 'n' terms of an AP, $S_n$ = $\frac{n}{2}[2a+(n-1)d]$,

where 'a' is the first term and 'd' is the common difference

Solution

Here, the given series is 21,18,15,12.... is 0

18 - 21 = -3

15 - 18 = -3

Since 18 - 21 = 15 -18 = -3  , the given series 21,18,15,12...... form an AP

The first term = a = 21

Common difference = d= -3

Given that,

sum to n-terms o f the series 21,18,15,12.... is 0

$S_n$ = 0

$\frac{n}{2}[2a+(n-1)d]$ = 0

Substituting the value of a = 21 and d = -3 we get

$\frac{n}{2}[2 X21+(n-1)X-3]$ =  0

$\frac{n}{2}[42 -3n +3] = 0$

$\frac{n}{2}[45 - 3n] = 0$

n(45 - 3n) = 0

n = 0, or 3n = 45

n = 0, n = 15

The number of terms such that the sum to n-terms is zero = 15

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