Question

If 5th term of the ap 22 and 14 th term of the ap is 58. Find the ap

Answer 1

Answer:

6,10,14,18,22.... where a=6 and d=4

Step-by-step explanation:

***Tn=a+(n-1)d***

So;

T5 =22=a+4d; a = 22-4d

(We can't derive the expression we got to the same equation of T5 so we derive it with equation of T14)

T14 =58=a+13d; 58=22d-4d+13d;= 22+9d

9d=36; So d=4 ( with this we can derive 'a')

We can use any equation for value of a;

T5 =a+4*4=22

=a+4*4=22a=22-16=6

So with this we can form an AP

= 6,10,14,18,22,26.....

Answer 2

Answer:

5th term of AP = 22

⇒ a + (n-1)d = 22

a + (5-1)d = 22

a + 4d = 22

a = 22 - 4d — (i)

14th term of AP = 58

⇒ a + (n-1)d = 58

a + (14-1)d = 58

a + 13d = 58

a = 58 - 13d — (ii)

Now, we compare equations (i) and (ii):-

Since we know that both the values of these equations stand for ‘a’, we can say that they are equal.

⇒ 22 - 4d = 58 - 13d

13d - 4d = 58 - 22

9d = 36

d = 4

Now, we find the value of ‘a’ by substituting the value of ‘d’ in equation (i):-

a = 22 - 4d

a = 22 - 4(4)

a = 22 - 16

a = 6

AP formed:-

6, 10, 14, 18, 22, 26, 30, 34, 38 .........

Hope it helps

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