Question

In an a.p 3,8,13....153 fixed 10th term from end

Answer 1

GIVEN :–

• An A.P. 3,8,13. . . . . . . . 153 .

TO FIND :–

• 10th term from end = ?

SOLUTION :–

• We know that –

$\\\longrightarrow \red {\bf{T_n=a+(n-1)d}}$

• Here –

⇛ a = 153 [From end]

⇛ d = -5

• Now put the values –

$\\\implies\bf T_{10}=153+(10-1)( - 5)$

$\\\implies\bf T_{10}=153+(9)( - 5)$

$\\\implies\bf T_{10}=153-45$

$\\\implies \large{\boxed{\bf T_{10}=108}}$

• Hence , 10th term from end of given A.P. is 108.

Answer 2

Answer:

Given :-

• In an A.P, 3 , 8 , 13 . . . . . . . . 153.

To Find :-

• What is the 10th term from the end.

Formula Used :-

${\red{\boxed{\large{\bold{t_n =\: a + d(n - 1)}}}}}$

where,

• a = First term
• d = Common difference
• n = Number of terms

Solution :-

Given :

• a = 153
• d = 3 - 8 = - 5
• n = 10

According to the question by using the formula we get,

${:}\implies$ $\sf t_{10} =\: 153 + (- 5)(10 - 1)$

${:}\implies$ $\sf t_{10} =\: 153 + (- 5)(9)$

${:}\implies$ $\sf t_{10} =\: 153 + (- 5) \times (9)$

${:}\implies$ $\sf t_{10} =\: 153 + (- 45)$

${:}\implies$ $\sf t_{10} =\: 153 - 45$

${:}\implies$ $\sf\bold{\purple{t_{10} =\: 108}}$

$\therefore$ The 10th term from the end is 108 .