Question

In an A.P,if a=1,tn=20 and Sn=462,then n is

Answer 1

Step-by-step explanation:

Given :-

In an AP , a = 1 , tn = 20 , Sn = 462

To find :-

The value of n

Solution :-

Given that

In an A.P. , a = 1

tn = 20

Sn = 462

We know that

Sum of the first "n" terms in an A.P.=

Sn = (n/2)(a+tn)

=> 462 = (n/2)(1+20)

=> 462 = (n/2)(21)

=> 462 = 21n/2

=> 462×2 = 21n

=> 924 = 21n

=> 21n = 924

=> n = 924/21

=> n = 44

Therefore, n = 44

Answer :-

The value of n is 44

Check :-

We have,

a = 1

tn = 20

n = 44

We know that

Sum of the first n terms in an A.P

= (n/2)[a+tn]

= (44/2)(1+20)

= 22×21

= 462

Verified the given relations in the given problem.

Used formulae:-

→ Sum of the first n terms in an A.P

= (n/2)[a+tn]

• a = First term
• tn = General or nth term
• n = Number of terms in the AP.

Answer 2

Given -

In an A.P,

• a = 1,
• tn = 20 and
• Sn = 462

To find -

• the value of n

Solution -

Recalling the formulas,

$\small{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: \: {\bf{S_n = \frac{n}{2}\{2a + (n - 1)d}\}}}$

$\small{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \star \: \: {\bf{t_n = a + (n -1)d}}}$

Here,

1. a = first term
2. d = common difference
3. Sn = Sum of nth terms
4. tn = nth term

Solving,

$\small{\sf{t_n = a + (n -1)d}}$

$\small{ \implies \: \: {\sf{20 -1 = (n -1)d}}}$

$\small{ \implies \: \: {\sf{(n -1)d=19}}}$

$\small{ \: \: {\sf{S_n = \frac{n}{2}\{2a + (n - 1)d}\}}}$

Putting (n -1)d = 19,

$\small{ \implies \: \: {\sf{ 462= \frac{n}{2}\{2(1) + 19\}}}}$

$\small{ \implies \: \: {\sf{ 462= \frac{n}{2}(2 + 19)}}}$

$\small{ \implies \: \: {\sf{ 462= \frac{n}{2}\times21}}}$

$\small{ \implies \: \: {\sf{ 462= \frac{21n}{2}}}}$

$\small{ \implies \: \: {\sf{ 462\times2= {21n}}}}$

$\small{ \implies \: \: {\sf{ 21n=924}}}$

$\small{ \implies \: \: {\sf{ n=\frac{\cancel{924}^{44}}{\cancel{21}}}}}$

$\small{ \implies \: \: {\bf{ n=44}}}$

Therefore, the value of n = 44.