Question

Find the sum of the first 19 terms of the Ap, whose nth term is given by tn=7-4n

Answer 1

an=7-4n
Put n=1
a1=7-4*1=3
Put n=2
a2=7-4*2=-1
d=-4
Sn=19/2[2*3+(19-1)-4]
=19/2(6-75)
=19/2*-66
=-627

Answer 2

The sum of the first 19 terms of the AP is -627.

Explanation:

Given : The nth term is given by $t_n=7-4n$

To find : The sum of the first 19 terms of the AP ?

Solution :

The nth term is given by $t_n=7-4n$

The first term, put n=1

$t_1=7-4(1)=3$

The second term, put n=2

$t_2=7-4(2)=-1$

The common difference is $d=t_2-t_1$

$d=-1-3=-4$

The sum of n number is given by,

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

The sum of 19 term, put n=19, a=3 and d=-4

$S_{19}=\frac{19}{2}[2(3)+(19-1)(-4)]$

$S_{19}=\frac{19}{2}[6+(18)(-4)]$

$S_{19}=\frac{19}{2}[6-72]$

$S_{19}=\frac{19}{2}\times (-66)$

$S_{19}=-627$

Therefore, the sum of the first 19 terms of the AP is -627.

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Find the sum of first 19 terms of the AP , whose nth term is given bt tn= 7-4n

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