Question

Which term of AP -20,-15,-10 is 100 more than 17th term​

Answer 1

Answer

• The 37th term will be 100 more than the 17th term

Explanation

Given

• AP is -20,-15,-20....

To Find

• The term which is 100 more then it's 17th term?

Solution

So here we shall first find the 17th term

→ aₙ = a+(n-1)d

• a = -20
• d = -15-(-20) = -15+20 = 5
• n = 17

→ a₁₇ = -20+(17-1)(5)

→ a₁₇ = -20+(16)(5)

→ a₁₇ = -20+80

→ a₁₇ = 60

✭ Term 100 more than the 17th term

→ 60+100 = 160

✭ Number of the term where it is 160

→ aₙ = a+(n-1)d

→ 160 = -20+(n-1)(5)

→ 160+20 = (n-1)(5)

→ 180 = (n-1)(5)

→ 180/5 = n-1

→ 36 = n-1

→ 36+1 = n

→ n = 37

Answer 2

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

▣Given Series is -20, -15, -10.........so, on

$\underline{\large{\purple{\bf{Find:-}}}}$

▣The term of the given series which will be 100 more than its 17th term.

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

F1stly we will find it's 17th term

So, we know that

$\huge{\underline{\boxed{\sf a_{n} = a + (n - 1)d}}}$

where,

• a = -20
• n = 17
• d = -15-(-20) = -15+20 = 5

So,

$\sf \dashrightarrow a_{17} = - 20 + (17- 1)(5)$

$\sf \dashrightarrow a_{17} = - 20 + (16)(5)$

$\sf \dashrightarrow a_{17} = - 20 + 80$

$\sf \dashrightarrow a_{17} = 60$

$\bold{\ast\underline{a_{17} = 60}}$

The no. which is 100 more than its 17th term

$\sf:\to a_{17} + 100$

$\sf:\to 60 + 100$

$\sf:\to 160$

Now, again using

$\huge{\underline{\boxed{\sf a_{n} = a + (n - 1)d}}}$

where,

• $\sf a_n = 160$
• a = -20
• d = 5

So,

$\sf \dashrightarrow 160= - 20 + (n- 1)(5)$

$\sf \dashrightarrow 160= - 20 + 5n- 5$

$\sf \dashrightarrow 160= - 20 - 5+ 5n$

$\sf \dashrightarrow 160= - 25+ 5n$

$\sf \dashrightarrow 160 + 25=5n$

$\sf \dashrightarrow 185=5n$

$\sf \dashrightarrow \dfrac{185}{5}=n$

$\sf \dashrightarrow 37=n$

$\sf \dashrightarrow n = {37}^{th} \: term$

Hence, 37th term is 100 more than its 17th term