Question

Find the 15th term from the end of the A.P 5,8,11,………92
pls ans this fast

Answer 1

Given:-

• AP = 5,8,11,.......92

• First term ( $\sf a_1$ ) = 5

• Last term ( $\sf a_n$ ) = 92

• Common difference = ( $\sf a_2- \sf a_1$ ) = 8 - 5 = 3

To find:-

• 15th term ( $\sf a_{15}$ ) = ?

Solution:-

$\bold{\underline{\boxed{\sf{\pink{\dag \; \sf a_{15} = a + 14d}}}}}$

★ Putting given values:-

$\implies \sf{ \sf a_{15} = 5 + 14 \times 3}$

$\implies \sf{ \sf a_{15} = 5 + 42}$

$\implies \sf{ \sf a_{15} = 47}$

$\bold{\underline{\underline{\sf{\pink{\dag \; Hence \; 15th \: term \; of \; AP = 47.}}}}}$

$\rule{200}{2}$

Answer 2

Answer:

a15= 47

Step-by-step explanation:

a1= 5                        (Given)

a2=8                        (Given)

an=92                      (As said in the question from the end term)

Common Difference (d)= a2-a1

                                      =8-5

                          thus d =3

Therefore a15 = a1 + 14d                            ( a1 or a is the same)

                        =5 + (14 x 3)                        ( ALWAYS MULTIPLY then add)  

                        =5 + 42

      Hence a15= 47

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