Question

The nth term of an A.P. is 5 – 2n. Find the A. P.
(a) 1, 3, 5..... (b) 3, 1, -1, -3..... (c) 3, 5, 7.....
(d) 4,2,0,-2​

Answer 1

Answer:

(b) 3, 1, -1, -3

Step-by-step explanation:

tn = 5-2n

t1 = 5-2×1 =5-2=3

t2= 5- 2×2 = 5-4 = 1

t3 = 5-2 ×3 = 5-6 = -1

t4 = 5-2× 4 = 5-8 = -3

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Answer 2

$\sf\huge\blue{\underline{\underline{ Question : }}}$

The nth term of an AP is 5 - 2n. Find the series of AP.

(a) 1,3,5....

(b) 3,1,-1,-3....

(c) 3,5,7.....

(d) 4,2,0,-2....

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• nth term of AP = 5 - 2n.

★ To find,

• Series of AP.

★ Let,

• Take n = 1

$\tt\:\implies 5 - 2(1)$

$\tt\:\implies 5 - 2$

$\tt\:\implies 3$

• Take n = 2

$\tt\:\implies 5 - 2(2)$

$\tt\:\implies 5 - 4$

$\tt\:\implies 1$

• Take n = 3

$\tt\:\implies 5 - 2(3)$

$\tt\:\implies 5 - 6$

$\tt\:\implies - 1$

Therefore, the AP series = 3,1,-1,...

★ Verification,

• AP = 3,1,-1.....

• a1 = 3
• a2 = 1

→ Common difference (d) = a2 - a1

$\tt\:\implies 1 - 3$

$\tt\:\implies - 2$

Now,

• a = 3
• d = - 2
• n = n
• an = ?

By using nth term of AP

• $\bf\green{ : \implies a_{n} = a + (n - 1)d }$

$\tt\:\implies a_{n} = 3 + (n - 1)(-2)$

$\tt\:\implies a_{n} = 3 - 2n + 2$

$\tt\:\implies a_{n} = 5 - 2n$

Hence, it was verified.

$\underline{\boxed{\bf{\purple{ \therefore The\:AP\:series = 3,1,-1....(b)}}}}\:\orange{\bigstar}$

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★ Important formulas in AP :

$\rm\red{:\implies a_{n} = a + (n - 1)d }$

$\rm\red{:\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] }$

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