Question -
Answer -
(i) Given:
AP: 24, 23 ¼, 22 ½, 21¾, … = 24, 93/4, 45/2, 87/4, …
Here, a1 =a = 24, a2 = 93/4
Common difference, d =a2 – a1 = 93/4 – 24
= (93 – 96)/4
= – 3/4
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
We know, an =a + (n – 1) d
an =24 + (n – 1) (-3/4)
= 24 – 3/4n + ¾
= (96+3)/4 – 3/4n
= 99/4 – 3/4n
Now we need to find,first negative term.
Put an <0
an =99/4 – 3/4n < 0
99/4 < 3/4n
3n > 99
n > 99/3
n > 33
Hence, 34th termis the first negative term of given AP.
(ii) Given:
AP: 12 + 8i, 11 + 6i,10 + 4i, …
Here, a1 =a = 12 + 8i, a2 = 11 + 6i
Common difference, d =a2 – a1
= 11 + 6i – (12 + 8i)
= 11 – 12 + 6i – 8i
= -1 – 2i
We know, an =a + (n – 1) d [where a is first term or a1 and d is commondifference and n is any natural number]
an =12 + 8i + (n – 1) -1 – 2i
= 12 + 8i – n – 2ni +1 + 2i
= 13 + 10i – n – 2ni
= (13 – n) + (10 – 2n)i
To find purely realterm of this A.P., imaginary part have to be zero
10 – 2n = 0
2n = 10
n = 10/2
= 5
Hence, 5th termis purely real.
To find purelyimaginary term of this A.P., real part have to be zero
∴ 13 – n = 0
n = 13
Hence, 13th termis purely imaginary.