Question

The 6th and 17th terms of an Arthemis progression are 19 and 41 respectively, find the 40th term No SPAM

Answer 1

Answer:

correct answer is 87

Step-by-step explanation:

if it is correct than mark brainliest

Answer 2

GIVEN :-

• 6th term of an AP is 19

• 17th term of an AP is 41

TO FIND :-

• 40th term of the AP

SOLUTION :-

In an AP the nth term is given by ,

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

Where ,

• a is first term

• d is common difference

Similarly ,

6th term of an AP is given by ,

$\implies \sf \: a_6 = a + (6 - 1)d \\ \\ \implies \large\bold {\boxed { \sf{ \bigstar | \:a_6 = a + 5d}}}$

We are given that 6th term of an AP = 19

$\implies \sf \: a + 5d= 19 \longrightarrow \: eq(1)$

In the same way 17th term of an AP is given by ,

$\large \bold {\boxed{ \sf{ \bigstar \: | a_{17} = a + 16d}}}$

We are given that 17th term of an AP = 41

$\implies \sf \: a + 16d= 41 \longrightarrow eq(2)$

Subtracting equation(1) from equation(2) ,

$\: \: \: \: \: \: \: \: \: \: \: \sf \: a + 16d = 41 \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: a \: + \: \: 5d = \: \: 19 \\ \: \: \: \: \: \: \: \: ( - ) \: \: ( - ) \: \: \: \: \: ( - )\\ \: \: \: \: \: \: \: \: \: \: - - - - - - - \\ \: \: \: \: \: \: \: \implies \sf \: 11d = 22 \\ \: \: \: \: \: \: \: \implies \sf \: d = \dfrac{22}{11} \\ \: \: \: \: \: \: \: \implies \sf {\underline {\boxed {\pink{ d = 2}}}}$

From eq(1) ,

$\implies \sf \: a + 5(2) = 19 \\ \\ \implies \sf \: a + 10 = 19 \\ \\ \implies \sf \: a = 19 - 10 \\ \\ \implies {\bold {\underline{\boxed{ \pink{ \sf{a = 9}}}}}}$

40th term of an AP is given by ,

$\large \bold {\boxed {\sf{ \bigstar | \:a_{40} = a + 39d}}}$

$\implies \sf \: a_{40} = 9 + 39(2) \\ \\ \implies \sf \: a_{40} = 9 + 78 \\ \\ \implies \sf \: a_{40} = 87$

∴ The 40th term of the given AP is 87

ADDITIONAL INFO :-

Sum upto nterms of an AP is given by ,

$\large\bold{\boxed{\sf{\bigstar|\:S_n\:=\:\dfrac{n}{2}[2a+(n-1)d]}}}$