Question

Q. If the common difference of an A.P. is 3, then find a20 – a15.​

Answer 1

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Sol.

Let the first term of the AP be a.

an = a(n − 1)d

a20  – a15 = [a + (20 – 1)d] – [a + (15 – 1)d]

              = 19d – 14d

              = 5d

              = 5 × 3

= 15

Answer 2

Given :

• Common difference of an AP is 3.

To find :

• t (20) - t (15)

Solution :

We know that,

$tn \: = a + (n - 1)d$

here,

$t(20) = a + 19d$

d = 3

$\implies \: t(20) = a + (19 \times 3)$

$\implies \: t(20) = a + 57$

Similarly,

$t(15) = a + (15 - 1)d$

$t(15) = a + 14d$

here, d = 3

$\implies \: t(15) = a + (14 \times 3)$

$\implies \: t(15) = a + 42$

Now we need to find their difference ,

$t(20) - t(15) = (a + 57) - (a + 42)$

$\implies \: a + 57 - a - 42$

$\implies \: (a - a) + 15$

$\implies \: 15.$

Therefore, the difference between a(20) and a(15) is 15.