Question

in an AP 4th and 7th terms are 17 and 23 respectively find 'd' and 'a'
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Answer 1

Answer:-

Given:

4th term of an AP = 17

7th term = 23

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

a + (4 - 1)d = 17

→ a + 3d = 17 -- equation (1)

Similarly,

a + (7 - 1)d = 23

→ a + 6d = 23 -- equation (2)

Subtract equation (1) from (2).

→ a + 6d - (a + 3d) = 23 - 17

→ a + 6d - a - 3d = 6

→ 3d = 6

→ d = 6/3

→ d = 2

Substitute the value of d in equation (1).

→ a + 3(2) = 17

→ a + 6 = 17

→ a = 17 - 6

→ a = 11

Therefore,

• First term is 11.

• common difference is 3.

Answer 2

Question : -

in an ap 4th and 7th term are 17 and 23 respectively Find D and A ?

Solution :-

Given that ;

• fourth term of an AP = 17 and 7th term = 23

What to find here ?

we have to find here the D and A respectively .

Now,

Nth term of an AP -a (n)=a+(n-1)d

So,

in the case 1

$a + (4 - 1)d = 17$

$= > a + 3d = 17$

Now again in ,

case 2

$a(7 - 1)d = 23$

$= > a + 6d = 23$

now we have to subtract the case one from the case 2 we get ;

$a + 6d - (a + 3d) = 23 - 17$

$= > a + 6d - a - 3d= 6$

$= > 3d = 6$

$d = \frac{6}{3}$

$d = 2$

Now,

according to the given question we have to put the value of D in the case of 1 .

$a + 3 \times 2 = 17$

$= > a + 6 = 17$

$= > a = 17 - 6$

$= > a = 11$

so the first term will be 11 and the common difference will be 3 .