Question

In an arithmetic sequence sum of 1st and 3rd term is 6 and sum of 2nd and 4th term is 20. find its 11th term?​

Answer 1

Given : -

Sum of 1st and 3rd term = 6

Sum of 2nd and 4th term = 20

Required to find : -

• 11th term ?

Formula used : -

$\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}$

Here,

• a = first term

• d = common difference

• n = term number

Solution : -

Sum of 1st and 3rd term = 6

Sum of 2nd and 4th term = 20

we need to find the 11th term of the sequence .

So,

1st term + 3rd term = 6

2nd term + 4th term = 20

However,

1st term can be represented as " a "

2nd term can be represented as " a + d "

3rd term can be represented as " a + 2d "

4th term can be represented as " a + 3d "

This implies ;

=> a + ( a + 2d ) = 6

=> a + a + 2d = 6

=> 2a + 2d = 6 $\purple{\longrightarrow{\tt{\pink{ Equation - 1 }}}}$

Consider this as equation 1

Similarly,

=> a + d + ( a + 3d ) = 20

=> a + d + a + 3d = 20

=> 2a + 4d = 20 $\purple{\longrightarrow{\tt{\pink{ Equation - 2 }}}}$

Consider this as equation 2

Now,

Let's try to solve these 2 equations simultaneously .

Using the elimination method let's try to eliminate one variable in order to simplify the calculations .

So,

Subtract Equation 1 from Equation 2

$\tt 2a + 4d = 20 \\ \tt 2a + 2d = 6 \\ \underline{ (-) (-) \: \: (-) \: \: \: \: }\\ \tt \underline{ \: \: \: \: \: \: \: \: 2d = 14 \: \: \: } \\ \\ \implies \tt 2d = 14 \\ \\ \implies \sf d = \dfrac{14}{2} \\ \\ \implies \sf d = 7$

Hence,

• Common difference ( d ) = 7

Substitute the value of d in equation 1

=> 2a + 2d = 6

=> 2a + 2 ( 7 ) = 6

=> 2a + 14 = 6

=> 2a = 6 - 14

=> 2a = - 8

=> a = - 8/2

=> a = - 4

Hence,

• First term ( a ) = - 4

Using the formula ;

$\boxed{\tt{ {a}_{nth} = a + ( n - 1 ) d }}$

This implies ;

$\to\sf{ {a}_{nth} = {a}_{11} }$

$\to\sf{ {a}_{11} = - 4 + ( 11 - 1 )7 }$

$\to\sf{ {a}_{11} = - 4 + ( 10 ) 7 }$

$\to\sf{ {a}_{11} = - 4 + 70 }$

$\to\sf{ {a}_{11} = 66 }$

Therefore,

11th term = 66

Additional Information : -

To find the sum of n terms of any given arithmetic sequence . we should use the formula ;

$\boxed{\rm{ {S}_{nth} = \dfrac{n}{2} [ 2a + ( n - 1 ) d ] }}$

QUESTION :-

How to identify whether the given sequence is an arithmetic progession or not ?

Answer :-

To find whether any given sequence is an arithmetic progession or not we need to use a small trick .

The small trick is ;

The common difference between the terms of the arithmetic progession should be equal / constant .

Common difference = ( 2nd term - 1st term ) = ( 3rd term - 2nd term )

If this condition get's satisfied then we can say that the given sequence is an arithmetic progession .

Answer 2

$\huge\tt{\underline{\underline{Given:-}}}$

• Sum of 1st and 3rd term of an AP is 6
• Sum of 2nd and 4th term of an AP is 20

$\huge\tt{\underline{\underline{To\:find}}}$

• 11th term of the AP

$\huge\tt{\underline{\underline{Solution:-}}}$

In an AP nᵗʰ term is given by,

$\large\bold{\underline{\boxed{a_n\:=\:a+(n-1)d}}}$

$\large\rm{a_n\rightarrow{nth\:term\:of\:an\:AP}}$

$\large\rm{a\rightarrow{First\:term\:of\:an\:AP}}$

$\large\rm{d\rightarrow{Common\:difference\:of\:an\:AP}}$

$\large\rm{n\rightarrow{The\:number\:of\:the\:particular\:term}}$

Then 3rd term Of an AP is given by,

$\large\rm{a_3\:=\:a+(3-1)d}$

$\large\rm{\implies{a_3\:=\:a+2d}}$

Given Sum of first and third term of an AP = 6

$\large\rm{\implies{6\:=\:a+a+2d}}$

$\large\rm{\implies{6\:=\:2a+2d\rightarrow\:Equation(1)}}$

$\large\rm{\implies{a_3\:=\:a+2d}}$

In the same way ,

Second term of an AP is

$\large\rm{a_2\:=\:a+(2-1)d}$

$\large\rm{\implies{a_2\:=\:a+d}}$

Fourth term of an AP is given by,

$\large\rm{a_4\:=\:a+(4-1)d}$

$\large\rm{\implies{a_4\:=\:a+3d}}$

Given Sum of 2nd and 4th term of an AP = 20

$\large\rm{\implies{a+d+a+3d\:=\:20}}$

$\large\rm{\implies{2a+4d\:=\:20\rightarrow\:Equation(2)}}$

Subtraction Equation(1) from Equation(2) ,

$\: \: \: \: \: 2a + 4d = 20 \\ \\ \: \: \: \: \: \: 2a + 2d = 6 \\ \\ \: \: ( - ) \: \: \: \: ( - ) \: \: \: \: ( - ) \\ \: \: - - - - - - - -$

$\large\rm{\implies{2d\:=\:14}}$

$\large\rm{\implies{d\:=\:7}}$

From Equation(1),

2a + 2d = 6

$\large\rm{\implies{2a+2(7)\:=\:6}}$

$\large\rm{\implies{2a\:=\:-8}}$

$\large\rm{\implies{a\:=\:-4}}$

Eleventh term in an AP is given by ,

$\large\rm{Eleventh\:term\:=\:a+(11-1)d}$

$\large\rm{\implies{Eleventh\:term\:=\:-4+10(7)}}$

$\large\rm{\implies{Eleventh\:term\:=\:66}}$

∴ The Eleventh term of the AP is 66

$\large\tt{\underline{\underline{\pink{Additional\:information:-}}}}$

❃ Sum upto n terms of an AP is given by ,

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

$\large\rm{S_n\rightarrow{Sum\:upto\:n\:terms}}$

$\large\rm{a\rightarrow{First\:term\:of\:an\:AP}}$

$\large\rm{d\rightarrow{Common\:difference\:of\:an\:AP}}$

$\large\rm{n\rightarrow{Last\:term\:of\:AP}}$