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NISA [10]
3 years ago
15

Plspls help!

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0

Answer:

a)=300

b)=79

Step-by-step explanation:

a) 3m²

=3(10)²

=300

b) n³/3 + 7m³/100

=3³/3 + 7(10)³

=9+70

=79

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AC # A'B'

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Question -
Sauron [17]

\rule{200}4

Answer : <em>The</em><em> </em><em>required</em><em> </em><em>ratio</em><em> </em><em>is</em><em> </em><em>(</em><em>1</em><em>4</em><em>m</em><em>-</em><em>6</em><em>)</em><em>:</em><em>(</em><em>8</em><em>m</em><em>+</em><em>2</em><em>3</em><em>)</em><em> </em><em>.</em>

\rule{200}4

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27}  \\

As , we know that the sum of n terms of an AP is given by ,

\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\

Assume that ,

  • First term of 1st AP = a
  • First term of 2nd AP = a'
  • Common difference of 1st AP = d
  • Common difference of 2nd AP = d'

Using this we have ,

\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\

\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\

Now also we know that the nth term of an AP is given by ,

\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\

Therefore,

\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\

\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\

From equation (i) and (iii) ,

\longrightarrow\sf\small n-1 = 2m-2\\

\longrightarrow\sf\small n = 2m -2+1 \\

\longrightarrow\sf\small n = 2m -1 \\

Substitute this value in equation (i) ,

\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\

Simplify,

\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\

\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\

\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\

From equation (ii) ,

\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\

\rule{200}4

8 0
3 years ago
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