Question

Find the 10th term of sequence whose 7th and 12th term are34and64 respectively

Answer 1

a+6d=34

a+11d=64

-5d=34-64

d=6

a=_2

a+9d=52

Answer 2

The value of 10 th term would be 52.

Step-by-step explanation:

Since we have given that

7th term = 34

12th term = 64

So, it becomes,

$a+6d=34\\\\a+11d=64\\\\----------------------------\\\\-5d=-30\\\\d=\dfrac{30}{5}\\\\d=6$

So, the value of a would be

$a+6d=34\\\\a+36=34\\\\a=34-36=-2$

Hence, the value of 10 th term would be

$a+9d=-2+9\times 6=-2+54=52$

# learn more:

Consider a sequence whose 3rd term is 13 and whose 7th term is 208. assume the sequence is geometric. find the function a(n) that describes the sequence and find the 19th term of the sequence

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