Q. 144.7( 6 Votes )

# In an A.P. a = 8, T_{n}= 33, S_{n} = 123, find d and n.

Answer :

We have a = 8

We know that, T_{n} = a + (n – 1)d

And from the question we can say that, T_{n}= 53

So we get,

⇒ 33 = a + (n – 1)d

Putting value of a in the above equation,

⇒ 33 = 8 + (n – 1)d

⇒ (n – 1)d = 33 – 8

⇒ (n – 1)d = 25

Now we have that, S_{n} = × (2a + (n – 1)d)

From the question we can say that, S_{n} = 123

So, we have,

⇒ 123 = × (2a + (n – 1)d)

⇒ 246 = n × (2a + (n – 1)d)

now we have a = 8 and (n – 1)d = 25, so we put them in the above equation,

⇒ 246 = n × (2(8) + 25)

⇒ 246 = n × (16 + 25)

⇒ 246 = n × (31)

⇒ n= = 6

⇒ n = 6

now, we know that, (n – 1)d = 25

we put the value of n in the above equation, we get

⇒ (6 – 1)d = 25

⇒ 5d = 25

⇒ d = 5

∴ for the given A.P, value of n is 6 and value of d is 5.

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