Question

find the common difference and three more terms of the given AP: 2,14,26,....​

Answer 1

Answer:

find the common difference and three more terms of the given AP: 2,14,26,....

Answer 2

Question

find the common difference and three more terms of the given AP: 2,14,26,....

Answer

Given:- 2,14,26,.... is an Arithmetic Progression

To find: the common difference and the more terms of the given AP.

Solution

We have,

$\tt \: a_{1} = 2$

$\tt \: a_{2} = 14$

$\tt \: a_{3} = 26$

$\rm \:◌ \: common \: difference = \tt \: a_{2} - a_{1}$

$\rm \: ◌ \:common \: difference = \tt \:14 - 2$

$\rm \: ◌ \: common \: difference(d) = \tt 12$

We know,

$\tt \: a _{n} = a _{1} + (n - 1)d$

Now,

Let's find the other terms of the AP .

The fourth↓

$\tt \: a _{4} = a _{1} + (4- 1)d$

$\leadsto \:\tt \: a _{4} = a _{1} +3d$

$\leadsto \:\tt \: a _{4} = 2 +3 \times 12$

$\leadsto \:\tt \: a _{4} = 38$

The fifth↓

$\tt \: a _{5} = a _{1} + (5- 1)d$

$\leadsto\tt \: a _{5} = a _{1} +4d$

$\leadsto\tt \: a _{5} = 2 +4 \times 12$

$\leadsto\tt \: a _{5} = 50$

The sixth↓

$\tt \: a _{6} = a _{1} + (6- 1)d$

$\leadsto \tt \: a _{6} = a _{1} + 5d$

$\leadsto \tt \: a _{6} = 2 + 5 \times 12$

$\leadsto \tt \: a _{6} = 62$

Hence the three more terms are 38,50 and 62.

There common difference is 12

Keep in mind

$\rm \:sum \: of \: terms = \frac{n}{2}(2a + (n - 1)d$

$\rm \: sum \: of \: natural \: nos. = \frac{n(n + 1)}{2}$