Question

The sum of first 6 terms of an ap is 36 and the sum of its first 16 term is 256 find the sum of first 10 terms of this AP

Answer 1

Question :-

The sum of first 6 terms of an ap is 36 and the sum of its first 16 term is 256 find the sum of first 10 terms of this AP.

Answer :-

• Sum is 100.

Step by step explanation :-

By first condition,

We know that,

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

Now,

$\rm:\longmapsto{36 = \frac{6}{2} [2a + (6 - 1)d]} \: \: \: \: \: \\ \\ \rm:\longmapsto{ \frac{72}{6} = 2a + 5d} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{2a + 5d = 12} - - - (1)$

by second condition,

$\rm:\longmapsto{256 = \frac{16}{2}[2a + (16 - 1)d] } \: \: \: \: \\ \\\rm:\longmapsto{ \frac{256 \times 2}{16} = 2a + 15d} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{2a + 15d = 32 - - - - (2)}$

Subtracting (1) from (2),

$\rm:\longmapsto{2a + 15d - (2a + 5d) = 32 - 12} \\ \\ \rm:\longmapsto{2a - 2a + 15d - 5d = 20} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{10d = 20} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{d = \cancel \frac{20}{10} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{d = 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Substituting d = 2 in Eqⁿ (1)

$\rm:\longmapsto{2a + 5(2) = 12} \\ \\ \rm:\longmapsto{2a + 10 = 12} \: \: \: \\ \\ \rm:\longmapsto{2a = 12 - 10} \: \: \\ \\ \rm:\longmapsto{2a = 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{a = 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\small\bf{\therefore \: The \: sum \: of \: first \: 10 \: terms \: of \: an \: AP}$

$\rm:\longmapsto{S_{10} = \frac{10}{2} [2(1) + (10 - 1)2]} \\ \\ \rm:\longmapsto{S_{10} = 5 (20) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{S_{10} = 100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer :

• The sum of first 10 terms is 100.

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Answer 2

Step-by-step explanation:

The sum of first 6 terms of an ap is 36 and the sum of its first 16 term is 256 find the sum of first 10 terms of this AP.

Answer :-

Sum is 100.

Step by step explanation :-

By first condition,

We know that,

\begin{gathered}\rm:\longmapsto{S_n = \frac{n}{2} [2a + (n - 1)d]} \\ \end{gathered}

:⟼S

n

=

2

n

[2a+(n−1)d]

Now,

\begin{gathered}\rm:\longmapsto{36 = \frac{6}{2} [2a + (6 - 1)d]} \: \: \: \: \: \\ \\ \rm:\longmapsto{ \frac{72}{6} = 2a + 5d} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{2a + 5d = 12} - - - (1)\end{gathered}

:⟼36=

2

6

[2a+(6−1)d]

:⟼

6

72

=2a+5d

:⟼2a+5d=12−−−(1)

by second condition,

\begin{gathered}\rm:\longmapsto{256 = \frac{16}{2}[2a + (16 - 1)d] } \: \: \: \: \\ \\\rm:\longmapsto{ \frac{256 \times 2}{16} = 2a + 15d} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{2a + 15d = 32 - - - - (2)}\end{gathered}

:⟼256=

2

16

[2a+(16−1)d]

:⟼

16

256×2

=2a+15d

:⟼2a+15d=32−−−−(2)

Subtracting (1) from (2),

\begin{gathered}\rm:\longmapsto{2a + 15d - (2a + 5d) = 32 - 12} \\ \\ \rm:\longmapsto{2a - 2a + 15d - 5d = 20} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{10d = 20} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm:\longmapsto{d = \cancel \frac{20}{10} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{d = 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

:⟼2a+15d−(2a+5d)=32−12

:⟼2a−2a+15d−5d=20

:⟼10d=20

:⟼d=

10

20

:⟼d=2

Substituting d = 2 in Eqⁿ (1)

\begin{gathered}\rm:\longmapsto{2a + 5(2) = 12} \\ \\ \rm:\longmapsto{2a + 10 = 12} \: \: \: \\ \\ \rm:\longmapsto{2a = 12 - 10} \: \: \\ \\ \rm:\longmapsto{2a = 2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{a = 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

:⟼2a+5(2)=12

:⟼2a+10=12

:⟼2a=12−10

:⟼2a=2

:⟼a=1

\small\bf{\therefore \: The \: sum \: of \: first \: 10 \: terms \: of \: an \: AP}∴Thesumoffirst10termsofanAP

\begin{gathered}\rm:\longmapsto{S_{10} = \frac{10}{2} [2(1) + (10 - 1)2]} \\ \\ \rm:\longmapsto{S_{10} = 5 (20) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto \red{S_{10} = 100} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}

:⟼S

10

=

2

10

[2(1)+(10−1)2]

:⟼S

10

=5(20)

:⟼S

10

=100

Answer :

The sum of first 10 terms is 100.

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The sum of 4th and 6th term of g.p is 80 while the product of 3rd and 5th term is 256 find the first term.

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