Question

7 times the 7th term of an ap is equals to 10 times the 10th term of an ap then show that 17th term is equals to zero​

Answer 1

Answer:

Step-by-step explanation:

7 x a7 = 10 x a10

7(a + 6d) = 10(a + 9d)

7a + 42d = 10a + 90d

3a = -48d

a = -16d

a17 = a + 16d = -16d + 16d = 0 (proved)

Answer 2

Proved as follows:

Step-by-step explanation:

• Any term in an AP is given by $a_n = a+(n-1)d$

• Thus, the seventh term would be

$a_7 = a+(7-1)d$

$a_7 = a+6d ----------------------(equation 1)$

• Similarly, the tenth term of the AP would be

$a_{10} = a+(10-1)d$

$a_{10} = a+9d ------------------(equation 2)$

• Given:

Seven times the seventh term = Ten times the tenth term

$7 \times a_7 = 10 \times a_{10$

Substituting the values from equations (1) and (2)

$7(a + 6d) = 10(a + 9d)$

$7a + 42d = 10a + 90d$

$7a -10a = 90 d - 42 d$

$-3a = 48d$

or

$3a = -48d$

$a = -16d-----------------------(equation 3)$

• Now, the seventeenth term would be

$a_{17} = a+(17-1)d$

$a_{17} = a+16d$

• Putting the value of $a$ from equation (3)

$a_{17} = a + 16d = -16d + 16d = 0$

Hence Proved.