Question

If the 10th term of an AP is 52 and 17th term is 20 more than the 13th term, find the AP
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Answer 1

Answer:

$a = 7 \\ d = 5$

AP: 7, 12, 17, 22, 27....

Step-by-step explanation:

We have formula of tn as:

$tn = a + (n - 1)d$

given:

$t10 = 52 \\ t17 = 20 + t13$

Now,

$t10 = a + 9d = 52 \\ t17 = 20 + (a + t12 + d)$

By formula of tn,

$t17 = a + 16d \\ t17 = (a + 12d) + (4d)$

So now we have:

$t17 = 20 + (a + 12d) = 4d + (a + 12d)$

Hence,

$20 = 4d \\ d = 5$

Now substituting value of d for t10 we get

$t10 = a + 9d = 52 \\ a + 9 \times 5 = 52 \\ a + 45 = 52 \\ a = 52 - 45 \\ a = 7$

Hence,

$a = 7 \\ d = 5$

Which gives:

AP: 7, 12, 17, 22, 27....

Answer 2

Refers to attachment ✔✔