Question

for an ap t6=-10 & t14=-34 find the value of t10​

Answer 1

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Answer 2

Answer:

The value of t10 is -22

Step-by-step explanation:

Let the first term  of the series be 'a'

Let the common difference be 'd'

For arithmetic progression , we know that nth term is a+(n-1)d

given that t6 = -10

$a+(6-1)d=-10\\ a+5d=-10............(1)$

Similarly t14= -34

$T14=a+13d=-34\\a+13d=-34..............(2)$

Subtracting (2) from (1) we get:-

$5d-13d=-10+34\\-8d=24\\d=-3$

substitute d value in equation 1, we get a

$a+5d=-10 \\a + 5(-3)=-10\\a-15=-10\\a=-10+15\\a= 5$

Hence required

$t10 = a+9d\\t10= 5+9(-3)\\t10= 5-27\\t10= -22$

The value of t10 is -22