Question

find the 6th term from end of the AP. 17,14,11,...(-40).​

Answer 1

ANSWER :–

-25

EXPLANATION :–

GIVEN :–

An A.P. ➩ 17 , 14 , 11 , ........... , (-40) .

TO FIND :–

Sixth term from end of given A.P.

SOLUTION :–

● First we have to find total term of A.P. –

☞ L = a + (n - 1)d

• Here , L = last term

a = first term

n = total term

d = common difference

➨ - 40 = 17 + (n - 1)(-3)

➨ - 40 - 17 = (n - 1)(-3)

➨ - 57 = (n - 1)(-3)

➨ n - 1 = 19

➨ n = 20

● Now we have to find 6 th term from end –

☞ 6th term from end = 15 term from beginning

➨ 15th term = a + 14 d

➨ 15th term = 17 + 14 (-3)

➨ 15th term = 17 - 42

➨ 15th term = -25

➨ Hence , 6 th term from end is -25.

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

• First term (a) = 17
• Common difference (d) = -3
• Last term (l) = -40

We know that , the First n terms of an AP is given by

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

Thus ,

➡-40 = 17 + (n - 1) × (-3)

➡-57 = (n - 1) × (-3)

➡n - 1 = 57/3

➡n - 1 = 19

➡n = 20

Hence , the total number of terms of an AP is 20

It can be observe that ,

The 6th term of an AP from end = 15th term of an AP from start

Thus ,

$\sf \mapsto a_{15} = 17 + (15 - 1) \times ( - 3) \\ \\\sf \mapsto a_{15} = 17 + (14) \times ( - 3) \\ \\ \sf \mapsto a_{15} = 17 - 42 \\ \\\sf \mapsto a_{15} = - 25$

Hence , the 6th term of an AP from end is -25