Question

the 4th and 7th term of an ap are 17 and 23 respectively find a 15

Answer 1

Hey there !!


↪ Given :-

→ 4th term = 17.

→ 7th term = 23.


↪ Find :- 15th term.


→ Solution :-


=> 4th term = 17.

=> a + 3d = 17..........(1).


And,

=> 7th term = 23.

=> a + 6d = 23.........(2).


▶ Substracte in equation (1) and (2), we get


a + 3d = 17.
a + 6d = 23.
(-)..(-).......(-)
_________

=> -3d = -6.

=> d = $\frac{ - 6}{ - 3}$

=> d = 2.


▶ Now, put the value of ‘d’ in equation (1), we get

=> a + 3d = 17.

=> a + 3(2) = 17.

=> a + 6 = 17.

=> a = 17 - 6.

=> a = 11.


▶ Now,

=> 15th term = a + 14d.

= 11 + 14(2).

= 11 + 28.

$\huge \boxed{ \boxed{ \bf = 39. }}$



✔✔ Hence, it is solved ✅✅.

____________________________________




$\huge \boxed{ \boxed{ \boxed{ \mathbb{THANKS}}}}$



$\huge \bf{ \# \mathcal{B}e \mathcal{B}rainly.}$

Answer 2

$\mathtt{\huge{\underline{SOLUTION:-}}}$


Firstly, Review some Points which helps in this Question;

$\mathtt{\bold{\underline{Tn=a+(n-1)d}}}$

Now,


Considered on Question!!!

T4=a+3d [Equation-1]


a+3d=17



Next!!!



T7=a+6d [Equation 2]



a+6d=23
_____________________

Solving Equation 1 and 2 By Elimination method!!

a+3d=17

a+6d=23

Multiply Equation 1 by 1 and Multiply Equation 2 by 1, We gets;

a+3d=17

a+6d=23

____________
-3d=-6

d=6/3

D=2

Substitute the Value of d in Equation 1

a+3d=17

a+3(2)=17

a+6=17

a=17-6

a=11


$\mathtt{\huge{\underline{ACCORDING\: TO\: THE\: Question:-}}}$

T15=??

T15=a+(n-1)d

T15=a+14d

T15=11+14(2)

T15=11+28

T15=39


$\mathtt{\huge{\underline{HENCE, \: T15=39 :-}}}$