Question

in an AP a=12/15 and d= 1/4 find the first three terms​

Answer 1

Given:

• First term = a = 12/15
• Common difference = 1/4

To find:

First 3 terms = ?

Solution:

In an A.P:

• First term = a
• Second term = (a + d)
• Third term = (a + 2d)

For finding second term:

(a + d)

$= \dfrac{12}{15} + \dfrac{1}{4}$

$= \dfrac{48 + 15}{60}$

$= \bf{ \dfrac{63}{60} }$

∴ Second term = 1.05 or 63/60

For finding the third term:

Third term

= a + 2d

$= \dfrac{12}{15} + 2 \bigg(\dfrac{1}{4} \bigg)$

$= \dfrac{4}{5} + \dfrac{1}{2}$

$= \dfrac{8 + 5}{10}$

$= \dfrac{13}{10}$

∴ Third term = 13/10 or 1.3

Answer 2

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Given:-

• a= 12/15
• d=1/4

To find:-

• First 3 terms

Solution:-

$\large \: \: \: \: \: \: \:\rightarrow\:a_1= a$

$\large \: \: \: \: \: \: \: \rightarrow \:a_2 = a + d$

$\large \: \: \: \: \: \: \: \rightarrow \:a_3 = a_2 + d$

Hence.

First term=12/15

second term=12/15 + 1/4 = 21/20

Third term= 63/60 + 1/4 = 13/10