Question

An arithmetic progression consists of 31 terms, p1, p2, p2 and so on. If
p1, p7, p13, p19, p25, and p31 add up to a total of 372, find the value of
p16.​

Answer 1

$\bf p_{16} = 62$

Step-by-step explanation:

Given:

• An arithmetic progression consists of 31 terms, p1, p2, p3 and so on.
• If p1, p7, p13, p19, p25, and p31 add up to a total of 372.

To find:

• Find the value of p16.

Solution:

Concept\Formula to be used:

First term of A.P. is 'a', it's common difference is 'd'.

General term of AP:$\bf a_n = a + (n - 1)d$

Step 1:

Write the given term.

$p_1 = a$

$p_7 = a + 6d$

$p_{13}= a + 12d$

$p_{19} = a + 18d$

$p_{25} = a + 24d$

$p_{31} = a + 30d$

Step 2:

Add all terms.

As addition of all these terms is 372.

So,

$a + a + 6d + a + 12d + a + 18d + a + 24d + a + 30d = 372$

or

$6a + 90d = 372$

or

$6(a + 15d) = 372$

or

$(a + 15d) = 62$

or

According to the formula of general term.

$p_{16} = 62$

Thus,

$\bf p_{16} = 62$

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