Question

Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:

	a	d	n	\(a_n\)
(i)	7	3	8	….
(iI)	-18	…	10	0
(iii)	…	-3	18	-5
(iv)	-18.9	2.5	…	3.6
(v)	3.5	0	105	…

Answer 1

(i) a = 7, d = 3, n = 8, \(a_n\) = ?
We know that,
For an A.P. \(a_n\) = a + (n − 1) d
= 7 + (8 − 1) \(^3\)
= 7 + (7) \(^3\)
= 7 + 2
= 28
Hence, \(a_n\)= 28

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(ii) Given that a = −18, n = 10, \(a_n\)= 0, d = ?
We know that, 
\(a_n\) = a + (n − 1) d
0 = − 18 + (10 − 1) d
18 = 9d
d = \(\frac{18}{9}\)= 2
Hence, common difference, d = 2

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(iii) Given that d = −3, n = 18, an = −5
We know that,
\(a_n\) = a + (n − 1) d 
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51 
a = 51 − 5 = 46
Hence, a = 46

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(iv) a = −18.9, d = 2.5, an = 3.6, n = ?
We know that,
\(a_n\) = a + (n − 1) d 
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5
(n-1) = \(\frac{22.5}{2.5}\)
n-1 = 9
n = 10
Hence, n = 10

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(v) a = 3.5, d = 0, n = 105, \(a_n\) = ?
We know that,
\(a_n\) = a + (n − 1) d 
\(a_n\) = 3.5 + (105 − 1) 0 
\(a_n\) = 3.5 + 104 × 0 
\(a_n\) = 3.5
Hence, \(a_n\) = 3.5