Question

In an a.p ; a5=-9 , a9=7 , then find a14

Answer 1

Given:-

$\tt \blacktriangleright a_{5}=-9 \\\\ \blacktriangleright a_{9}=7$

To find:-

$\tt \blacktriangleright a_{14}$

Solution:-

ATQ,

$\tt a_{5}= -9 = a+4d\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow(1)$

$\tt a_{9}= 7 = a+8d\rightarrow\rightarrow\rightarrow\rightarrow\rightarrow(2)$

Solving, equation 1 and 2

d = 4

a = -25

Finding a14

$\tt a_{14}= a+13d\\\\=-25+13(4)\\\\=-25+52\\\\=27$

So, a14(14th term) is 27.

Answer 2

$\bf\large{\underline{Question:-}}$

In an a.p ; a5=-9 , a9=7 , then find a14

$\bf\large{\underline{Given :-}}$

• $\tt→ a_5= -9\\\tt→ a_9=7$

$\bf\large{\underline{To\: find:-}}$

• $\tt→ a_{14}=?$

$\bf\large{\underline{Solution:-}}$

$Let\\\tt→ a_5=-9 = a+4d----equ(1)\\\tt→ a_4=7= a+8d---equ(2)$

On solving both of the equation

we get,

★ d ( common difference) =4

★ a = -25

Now,

$\tt→ a_{14} = a + (14-1)d\\\tt→ a_{14}=-25+(13)×4\\\tt→ a_{14}=-25+52\\\tt→ a{14}=27$

Hence,

$\tt\huge{\pink{\underline{\red{\fbox{\purple{a_{14}=27}}}}}}$