Question

If sum of 3 and 8 terms of an A.P. is 7 and sum of 7 and 14 terms is -3 then find the 10 term.​

Answer 1

Answer:

Step-by-step explanation:

Given that sum of 3rd and 8th terms of an AP is 7

from general equation of nth term of AP is

$t_{n} =a+(n-1)d$

so 3rd term is

$t_{3} =a+2d$ and 8th term is $t_{8} =a+7d$

sum of these two is $t_{3} +t_{8} =2a+9d$=7

Also given that sum of 7th and 14th terms is -3

$t_{7} = a+6d\\t_{14} =a+13d$

if we add them it gives 2a+19d=-3

from above equation 2a+9d=7

so 2a+9d+10d=-3

7+10d=-3

10d=-10

so d=-1

substitute d in any the equations

2a+9d=7

2a-9=7

2a=16

a=8

required is 10th term so

a+9d=8+9(-1)=-1

10th term is -1

Answer 2

Answer:

a=6

d(common difference)=4

Tn = a + (n - 1) d, where Tn = nth term and a = first term.

102= 6 + (n-1)4

96=(n-1)4

24=n-1

n=25

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