Question

in an Arithmetic proggression 6th term is more than twice the third time the sum of 4th and 5th term is 5 times the second term find the 10th term of Arithmetic proggression​

Answer 1

Answer :-

$\;\\\large{\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}}$

Here the concept of Arithmetic Progression has been used. We see we are given the conditions of different terms and we are unknown to the first term and the common difference. We can take them as variable quantities and find the answer.

Let's do it !!

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★ Equations Used :-

$\:\\\large{\boxed{\sf{a_{n}\;=\;\bf{a\;+\;(n\:-\:1)d}}}}$

$\:\\\large{\boxed{\sf{a_{6}\;=\;\bf{1\;+\;2(a_{3})}}}}$

$\:\\\large{\boxed{\sf{a_{4}\;+\;a_{5}\;=\;\bf{5(a_{2})}}}}$

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★ Correct Question :-

In an Arithmetic progression 6th term is one more than twice the third term. The sum of 4th and 5th term is 5 times the second term. Find the 10th term of Arithmetic progression.

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★ Solution :-

Given,

» 6th term of A.P. = 1 + 2(3rd term of A.P.)

» (4th term + 5th term) of A.P. = 5 × 2nd term

• Let the first term of the given A.P. be 'a' and the common difference be 'd'

Then,

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~ Case I :-

Its that 6th of A.P. is one more than twice the 3rd term.

So,

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\; a_{6}\;=\;\bf{1\;+\;2(a_{3})}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;a\;+\;5d\;=\;\bf{1\;+\;2(a\;+\;2d)}}}$

(Since, an = a + (n -1)d)

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;a\;+\;5d\;=\;\bf{1\;+\;2a\;+\;4d}}}$

By transposing the like terms to the other side, we get,

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;a\;+\;5d\;-\;2a\;-\;4d\;=\;\bf{1}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;d\;-\;a\;\;=\;\bf{1}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;d\;\;=\;\bf{1\;+\;a}}}$

Let this be equation number (i).

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~ Case II :-

Its clear here that sum of 4th and 5th term is five times the 2nd term.

So,

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\; a_{4}\;+\;a_{5}\;=\;\bf{5(a_{2})}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\; (a\;+\;3d)\;+\;(a\;+\;4d)\;=\;\bf{5(a\;+\;d)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\; a\;+\;3d\;+\;a\;+\;4d\;=\;\bf{5a\;+\;5d}}}$

Transposing the like terms to the other side, we get,

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;5a\;-\;2a\;=\;\bf{7d\;-\;5d}}}$

$\\\:\;\;\;\;\;\large{\sf{:\rightarrow\;\;\;3a\;=\;\bf{2d}}}$

Let this be equation number (ii).

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~ For value of a and d :-

From equation (i) and equation (ii), we get,

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;3a\;=\;\bf{2(1\;+\;a)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;3a\;=\;\bf{2\;+\;2a}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;3a\;-\;2a\;=\;\bf{2}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;a\;=\;\bf{2}}}$

$\\\;\large{\boxed{\boxed{\tt{First\;\;term\;\;of\;\;A.P.,\;\;a\;\;=\;\;\bf{2}}}}}$

Now using equation (i) and value of a, we get,

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;d\;\;=\;\bf{1\;+\;a}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;d\;\;=\;\bf{1\;+\;(2)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;d\;\;=\;\bf{1\;+\;2}}}$

$\\\:\;\;\;\;\;\large{\sf{:\Longrightarrow\;\;\;d\;\;=\;\bf{3}}}$

$\\\;\large{\boxed{\boxed{\tt{Common\;\;Difference\;\;of\;\;A.P.,\;\;d\;\;=\;\;\bf{3}}}}}$

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~ For 10th term of A.P. :-

Now we got the values as -

• a = 4

• d = 3

• n = 10 (since here we need to find 10th term)

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{n}\;=\;\bf{a\;+\;(n\:-\:1)d}}}$

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{10}\;=\;\bf{(2)\;+\;(10\:-\:1)(3)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{10}\;=\;\bf{(2)\;+\;(9)(3)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{10}\;=\;\bf{(2)\;+\;(27)}}}$

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{10}\;=\;\bf{2\;+\;27}}}$

$\\\:\;\;\;\;\;\large{\sf{:\mapsto\;\;\;a_{10}\;=\;\bf{29}}}$

$\\\:\large{\underline{\underline{\rm{Thus,\;10^{th}\;term\;of\;A.P.\;is\;\;\boxed{\bf{29}}}}}}$

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★ Verification of the Answer :-

For verification, we need to simply apply the values we got into the equations we formed. Then,

~ Case I :-

$\\\:\;\sf{:\Rightarrow\;\;\;d\;\;=\;1\;+\;a}$

$\\\:\;\sf{:\Rightarrow\;\;\;(3)\;\;=\;1\;+\;(2)}$

$\\\:\;\sf{:\Rightarrow\;\;\;(3)\;=\;(3)}$

Clearly, LHS = RHS.

~ Case II :-

$\\\:\;\sf{:\Rightarrow\;\;\;3(2)\;=\;2(3)}$

$\\\:\;\sf{:\Rightarrow\;\;\;(6)\;=\;(6)}$

Clearly, LHS = RHS.

Here both the conditions are satisfied, so our answer is correct.

Hence, Verified.

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Answer 2

Step-by-step explanation:

formula : an = a + (n-1)d

6th term is more than twice the third term

a6 = a + (6-1)d

a6 = a + 5d ..........( 1 )

a3 = a + (3-1)d

a3 = a + 2d ............( 2 )

from equation 1 and 2

a6 = 2( a3 ) + 1

a + 5d = 2( a + 2d ) + 1

a + 5d = 2a + 4d + 1

5d - 4d = 2a - a + 1

d = a + 1 .......( 3 )

Sum of 4th and 5th term is 5 times the second term

4th term :

a4 = a + ( 4-1 )d

a4 = a + 3d .........( 4 )

5th term :

a5 = a + ( 5-1 )d

a5 = a + 4d ............( 5 )

2nd term :

a2 = a + ( 2-1 )d

a2 = a + d .............( 6 )

4th term + 5th term = 5 ( 2nd term )

( a + 3d ) + ( a + 4d ) = 5 ( a + d )

a + 3d + a + 4d = 5a + 5d

2a + 7d = 5a + 5d

7d - 5d = 5a - 2a

2d = 3a ...........( 7 )

2 ( a + 1 ) = 3a ( put equation 3 in equation 7 )

2a + 2 = 3a

2a - 3a = -2

-a = -2

a = 2 .......( 8 )

Put a = 2 in equation 7

2d = 3a

2d = 3×2

2d = 6

d = 6÷2

d = 3 .........( 9 )

Therefore. a = 2.

d = 3.

10th term :

a10 = a + ( 10-1 )d

a10 = a + 9d ...........( 10 )

Put equation 8 and equation 9 in equation 10

a10 = 2 + 9×3

a10 = 2 + 27

a10 = 29