Question

if 3ns &7th terms of an ap are 15 & 39 respectively find ap
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Answer 1

Given :-

3rd & 7th term of an AP are 15 & 39

Required to find :-

• Arithmetic Progression ?

Solution :-

Given data :-

3rd & 7th term of an AP are 15 & 39

we need to find the arithmetic progression .

So,

3rd term = 15

7th term = 39

But,

we know that ;

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

Similarly,

7th term can be represented as ; a + 6d

This implies ;

a + 2d = 15 $\longrightarrow{\tt{Equation-1}}$

consider this as equation - 1

a + 6d = 39 $\longrightarrow{\tt{Equation-2}}$

Now,

Let's solve these 2 equations simultaneously using the Elimination Method

Subtract equation 1 from equation 2

$\tt{a + 6d = 39} \\ \tt{a + 2d = 15} \\ \tt{ \underline{ (-) ( - ) \: ( - ) \: \: }} \\ \underline{ \: \: \: \: \: \: \: \: \tt{4d = 24} \: } \\ \\ \sf \implies 4d = 24 \\ \sf \implies d = \frac{24}{4} \\ \sf \implies d = 6$

Hence,

• Common difference ( d ) = 6

Substitute the value of d in equation - 1

a + 2d = 15

a + 2 ( 6 ) = 15

a + 12 = 15

a = 15 - 12

a = 3

Hence,

First term ( a ) = 3

Now,

Let's find the AP using the above values ;

$\rightarrowtail \sf 1st \: term = a = \boxed{3} \\ \\ \rightarrowtail \sf 2nd \: term = a + d = 3 + 6 = \boxed{9} \\ \\ \rightarrowtail \sf 3rd \: term = a + 2d = 3 + 2 \times 6 = 3 + 12 = \boxed{ 15} \\ \\ \rightarrowtail \sf 4th \: term = a + 3d = 3 + 3 \times 6 = 3 + 18 = \boxed{21}$

Therefore ,

AP = 3 , 9 , 15 , 21 . . . . . .

Additional information :-

To find the nth term of any given AP we need to use the formula ;

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

To find the sum of terms of any given AP we need to use the formula

$\leadsto\boxed{ \rm{ \red{ {s}_{nth} = \frac{n}{2} \big[2a + ( n - 1 ) d \big]}}}$

Answer 2

Given ,

• The third and seventh term of an AP are 15 and 39

We know that , the nth term of an AP is given by

$\rm \large \fbox{ a_{n} = a + (n - 1)d }$

Thus ,

a + (3 - 1)d = 15

a + 2d = 15 --- (i)

And

a + (7 - 1)d = 39

a + 6d = 39 --- (ii)

Subtract eq (i) from eq (ii) , we get

4d = 24

d = 6

Put the value of d = 6 in eq (i) , we get

a + 2(6) = 15

a + 12 = 15

a = 3

Now , the sequence of an AP is given by

a , a + d , a + 2d , ... , a + (n - 1)d

$\sf \therefore \underline{The \: AP \: is \: 3 \: , \: 9 \: , \: 15 \: , .....}$