Question

An
a.p. consists of 21 terms. the sum of the three terms in the middle is 129 and the sum of the last three terms is 237. find the first term and the common difference.

Answer 1

Dear Student,

Answer:

First term a =3
common difference d= 4.

Solution:

In an AP, nth term is given as Tn = a+ (n-1)d

n = 21

sum of last three terms i.e. T19, T20, T21

$a + 18d + a + 19d + a + 20d = 237 \\ \\ 3a + 57d = 237 \\ \\ a + 19d = 79 \: \: \: \: \: \: eq1$
sum of three middle terms T10,T11 and T12

$a + 9d + a + 10d + a + 11d = 129 \\ \\ 3a + 30d = 129 \\ \\ a + 10d = 43 \: \: \: \: \: \: eq2$


Subtract both equations
$a + 19d - a - 10d = 79- 43 \\ \\ 9d =36 \\ \\ d = \frac{36}{9} \\ \\ d = 4$

Put the value of d, into eq 1 or eq2

$a + 10d = 43 \\ \\ a + 40 = 43 \\ \\ a = 43 - 40 \\ \\ a = 3$


So ,First term a =3
common difference d= 4.

Answer 2

Given total number of terms in an AP = 21.

Given that the sum of three terms in the middle is 129.

= > T10 + T11 + T12 = 129

= > a + 9d + a + 10d + a + 11d = 129

= > 3a + 30d = 129

= > a + 10d = 43 ----- (1)

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Given that sum of last three terms is 237.

= > T19 + T20 + T21 = 237

= > a + 18d + a + 19d + a + 20d = 237

= > 3a + 57d = 237

= > a + 19d = 79 ----- (2)

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On solving (1) & (2), we get

= > a + 10d = 43

= > a + 19d = 79

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-9d = -36

d = 4.

substitute d = 4 in (1), we get

= > a + 10d = 43

= > a + 10(4) = 43

= > a + 40 = 43

= > a = 43 - 40

= > a = 3.

Therefore, The first term is 3 and Common difference = 4.

The required AP is 3,7,11....

Hope this helps!