Question

In an A.P., the first term is -1 and eleven times the sum of the first three terms is equal to twice the sum of the next three terms. Find the common difference of the A.P.

Answer 1

Solution :-

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as ;

T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as ;

d = T(n) - T(n-1)

• If the number of terms in an AP is n ( where n is odd ) ,then there will be a single middle term.

Also, [(n+1)/2]th term will be its middle term.

• If the number of terms in an AP is n ( where n is even ) ,then there will be two middle terms.

Also, (n/2)th and (n/2 + 1)th terms will be its middle terms.

• The sum up to nth terms of an AP is given as ;

S(n) = (n/2)•[2a + (n-1)•d] where a is the first term and d is the common difference.

• The nth term of an AP is also given as ;

T(n) = S(n) - S(n-1)

we have,

→ First term = a = (-1)

→ Let common Difference = d

So,

→ 11 * (sum of first three terms) = 2 * (sum of next three terms)

→ 11 * (a₁ + a₂ + a₃) = 2 * (a₄ + a₅ + a₆)

→ 11 * {a + (a + d) + (a + 2d)} = 2 * {(a + 3d) + (a + 4d) + (a + 5d)}

→ 11 * (3a + 3d) = 2 * (3a + 12d)

→ 11 * 3 * (a + d) = 2 * 3 * (a + 4d)

→ 11(a + d) = 2(a + 4d)

→ 11a + 11d = 2a + 8d

→ 11a - 2a = 8d - 11d

→ 9a = (-3)d

→ 9(-1) = (-3)d

→ d = (-9)/(-3) = 3 (Ans.)

Hence, the common difference of the A.P. is 3.

Answer 2

Step-by-step explanation:

Given:In an A.P., the first term is -1 and eleven times the sum of the first three terms is equal to twice the sum of the next three terms.

To find: Find the common difference of the A.P.

Solution:

First term of A.P. : a=-1

Common difference:d

Sum of first n terms=

$S_n = \frac{n}{2} [2a + (n - 1)d]$

Sum of first three terms

$S_3 = \frac{3}{2} [2( - 1) + 2d] \\ \\ S_3 = 3( d - 1) \: \: \: ...eq1$

Now,

Sum of next three terms can be calculated using difference of sum of 3 terms from six terms

$S_6 = \frac{6}{2} ( - 2 + 5d) \\ \\S_6 = 3(5d - 2) \\ \\ S_{4 - 6} = S_6 - S_3 \\ \\ = 3(5d - 2) - 3(d - 1) \\ \\ = 15d - 6 - 3d + 3 \\ \\S_{4 - 6} = 12d - 3 \: \: \: ...eq2$

ATQ

eleven times the sum of the first three terms is equal to twice the sum of the next three terms

$11S_3 = 2S_{4 - 6}$

place the value from eq1 and eq2

$11 \times 3(d - 1) = 2 \times (12d - 3) \\ \\ 33d - 33 = 24d - 6 \\ \\ 33d - 24d = - 6 + 33\\ \\ 9d = 27 \\ \\ d = \frac{27}{9} \\ \\ d=3$

Common difference is 3.

Hope it helps you.

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