Hello friend!! Apne last lesson me humne aapko Arithmetic Progression ka basic concept aur isske 10 basic question ko solve karna sikhaya tha. Hume puri ummid hai ki, aapko ekdam sahi se samjh aaya hoga. Agar aapne abhi tak iss article ko nahi padha toh nichey humane diya aap yaha se padh kar sikh lijiye.

❒ Arithmetic progression | Samantar shreni Basic concept | 10 Question with Solution Hindi

Aaj hum aapko arithmetic progression ka "n waa pad" (nth term) gyaat karna sikhane waale hai. Aaj hum sikhenge ki, "samantar shreni ka nth term kya hota hai?" Aur "A.P. ka nth term kaise gyaat karte hai?"

Aaj ke lesson me jo hum sikhenge wo kuch iss prakar se nichey diya gaya hai -:

• Samantar Shreni ka "n waa pad (nth term)" kya hota hai?
• Samantar shreni ka "n waa pad" ka formula kya hai?
• Samatar shreni ka "n waa pad (nth term)" kaise gyaat karte hai?
• "nth term" se related 10 Question aur Solution

• Samantar shreni ka "n waa pad" (nth term) kya hota hai?

Kisi Samantar shreni ka "n waa pad" (nth term) uss samantar shreni ka koi bhi "ek pad" yaa "ek number" hota hai.

Jaise-: Yadi kisi Samantar shreni 3, 5, 7, . . . me first term = 3, second term = 5, third term = 7, issi tarah se fourth term, 85th term, 102th term . . . . . ki kuch value hai.

Issi prakar agar hum ek general term nikalana ho toh usko hum nth term se prakat karte hai. Jaha "n" ek positive integer (dhanatmak purnank) hota hai. Means "n" = 1, 2, 3, 4, . . . . hota hai.

Jaise-: 4, 8, 12, 16, . . . samantar shreni hai. Jisme "1st term = 4",

"2nd term = 8"

"3rd term = 12"

. . . . . . . . . . . . .

Agar iss shreni me hume 26th term ki value gyaat karni ho toh yahi 26th term, nth term kahalayega. Yaha par -:

nth term = 26th term

26th term gyaat karne ke liye hum issi nth term ka formula use karenge.

Hume ummid hai ki, ab aapko "nth term kya hota hai? Ye achhe se samjh me aa gaya hoga. Ab hum isska formula dekh lete hai.

**• Samantar shreni ka "n waa pad (nth term)" ka formula kya hai?**

Samantar shreni ka "nth term" gyaat karne formula iss prakar hai :

a_n = a + (n-1) × d

Yaha par

a_n = n waa pad (nth term)

a = pratham pad (first term)

n = pado ki sankhya (number of term)

d = Saarv antar (common difference)

• A.P. ka nth term ka formula kaise prapt (derive) karte hai?

A.P. ka formula aap upar dekh chuke hai. Ab hum sikhenge ki, iss formula ka derive kaise karte hai? Yaa ye formula kaise aaya.

Maana a₁, a₂, a₃, . . . ek Samantar shreni hai, jisme pratham pad = a aur saarv antar = d hai.

Tab 1st pad a₁ = a

2nd pad a₂ = a + d

3rd pad a₃ = a₂ + d

= (a + d) + d

= a + 2d

4th pad a₄ = a₃ + d

= (a + 2d) + d

= a + 3d

Aap upar ke 2nd, 3rd, aur 4th pad ko dekhiye. Aapko kya pattern dikh raha hai?

Aap dekh sakte hai ki, 2nd pad gyaat karne ke liye hum (2-1)×d ko pratham pad (1st pad) me jodate (add) karte hai. Means,

2nd pad a₂ = a + d = a + (2-1)d

Issi prakar se 3rd pad gyaat karne ke liye hum (3-1)×d ko pratham pad (1st pad) me jodate (add) karte hai. Means,

3rd pad a₃ = a + 2d = a + (3-1)d

Issi prakar 4th pad ke liye hum likh sakte hai :

4th pad a₄ = a + 3d = a + (4-1)d

Toh issi prakar se hume nth pad ko gyaat karne ke liye (n-1)×d ko 1st pad me add karna hoga. Means,

a_n = a + (n-1)×d

• "nth term" se related 10 Question aur Solution

Hume ummid hai ki, ab aap Samantar shreni ka nth term (n waa pad) kya hota hai? Isska formula aur isska concept achhe se samjh aa gaya hoga. Ab hum isske question ko solve karna sikh lete hai.

Question-1. A.P. 2, 4, 6, 8, . . . ka 8 waa pad gyaat kijiye.

Solution-: A.P. 2, 4, 6, 8, . . . me a = 2 aur d = 4-2 = 2

Formula a_n = a + (n-1)×d ka use karne par a₈ = 2+(8-1)×2

= 2+7×2

= 2+14

= 16

Isliye, di huyi shreni ka 8 waa pad 16 hai. Ans.

Question-2. A. P. -4, -1, 2, 5, . . . ka 10th term (10 waa pad) gyaat kijiye.

Solution-: Di gayi A.P. me a = -4 aur d = -1-(-4) = -1+4 = 3

Formula a_n = a + (n-1)×d ka use karne par a₁₀ = -4+(10-1)×3

= -4+9×3

= -4+27

= 23

Isliye, di huyi shreni ka 10 waa pad 23 hai. Ans.

Question-3. A. P. -6, -11, -16, -21, . . . ka 34th term (34 waa pad) gyaat kijiye.

Solution-: A. P. -6, -11, -16, -21, . . . me a = -6 aur d = -11-(-6) = -11+6 = -5

Formula a_n = a + (n-1)×d ka use karne par a₃₄ = -6+(34-1)×(-5)

= -6+33×(-5)

= -6+(-165)

= -6-165

= -171

Isliye, di huyi shreni ka 34 waa pad -171 hai. Ans.

Question-4. A.P. 12, 18, 24, . . . ka kon-sa pad 96 hoga?

Solution-: A.P. 12, 18, 24, . . . me a = 12 aur d = 18-12 = 6.

Maana di huyi shreni ka "n" waa pad 96 hai. Tab, a_n = 96

Hum jaante hai ki, a_n = a + (n-1)×d

∴ 96 = a+(n-1)×d

96 = 12+(n-1)×6

96-12 = (n-1)×6

84 = (n-1)×6

84/6 = n-1

14 = n-1

n = 14+1

n = 15

Isliye di huyi A.P. ka 15 waa pad 96 hoga. Ans.

Question-5. Ek A.P. ka third term 5 aur seventh term 9 hai, toh A.P. gyaat kijiye.

Solution-: Maana A.P. ka first term = a aur common difference = d hai.

Diya hai A.P. ka third term = 5

So, a+(3-1)×d = 5

a+2d = 5 . . . . . . . .(i)

aur A.P. ka seventh term = 9

So, a+(7-1)×d = 9

a+6d = 9 . . . . . . . . .(ii)

Equation (ii) me se Equation (i) ko subtract karne par

a + 6d = 9

a + 2d = 5

- - -

4d = 4

d = 4/4

d = 1

Equation (i) me d = 1 rakhane par

a+2×1 = 5

a+2 = 5

a = 5-2

a = 3

Isliye, A.P. 3, 4, 5, 6, . . . hai. Ans.

Question-6. Do anko (two digit) ke kitne numbers 4 se vibhajy (divisible) honge.

Solution-: Hum jaante hai ki, two digit ke 12, 16, 20, . . . 96 numbers 4 se vibhajy hai.

Maana 12, 16, 20, . . . 96 tak total "n" numbers hai jo 4 se vibhajy hai.

Tab, n waa pad = 96

Or a+(n-1)×d = 96

Yaha a =12 aur d =4

Tab 12+(n-1)×4 = 96

(n-1)×4 = 96-12

(n-1)×4 = 84

(n-1) = 84/4

n-1 = 21

n = 21+1

n = 22

Isliye, two digit ke 22 numbers 4 se vibhajy honge. Ans.

Question-7. A. P. 18, 15 1 , 13, . . . -47

me kitne pad (terms) hai ?

Solution-: Di gayi A.P. me a = 18 aur d = 15 1 - 18

= 31/2 - 18

= (31-36)/2

= -5/2

Maana di gayi A.P. me total "n" term hai to, nth term = a_n = -47

⇒ a+(n-1)×d = -47

⇒ 18+(n-1)×(-5/2) = -47

⇒ (n-1)×(-5/2) = -47-18

⇒ (n-1)×(-5/2) = -65

⇒ (n-1) = -65×(-2/5)

⇒ n-1 = 13×2

⇒ n-1 = 26

⇒ n = 26+1

⇒ n = 27

Isliye, Di gayi A.P. me total 27 term hai. Ans.

Question-8. Samantar shreni 3, 8, 13, . . . 253 me last pad (term) se 20th term kya hoga ?

Solution-: Samantar shreni 3, 8, 13, . . . 253 me last pad (term) 273 se a = 253 aur d = -5

∴ A.P. ka last pad se 20th term = a₂₀

a₂₀ = a+(n-1)×d

= 253+(20-1)×(-5)

= 253+19×(-5)

= 253-95

= 158

Isliye, di gayi A.P. ka last pad (term) se 20th term se 158 hai. Ans.

Question-9. Kya Samantar shreni 11, 8, 5, 2, . . . ka ek pad (term) -150 hai? Agar nhi toh Kyu?

Solution-: Di gayi Samantar shreni 11, 8, 5, 2, . . . me a = 11 aur d = -3

Maana iss A.P. ka nth term -150 hai. Hum jaanate hai ki, nth term = a_n = a+(n-1)d

⇒ -150 = 11+(n-1)×(-3)

⇒ -150 = 11-3n+3

⇒ -150 = 14-3n

⇒ -150 -14 = -3n

⇒ -164 = -3n

⇒ n = 164/3

"n" ek positive number (dhanatmak purnank) nahi hai. Isliye -150 iss A.P. ka pad (term) nahi hai. Ans.

Question-10. Kisi A.P. ke 4th aur 8th term ka yogphal 24 hai tatha 6th aur 10th term ka yogphal 44 hai. Iss Samantar shreni ka 1st, 2nd aur 3rd term gyaat kijiye.

Solution-: Maana Samantar shreni ka 1st term = a aur common difference = d hai.

∴ 4th term = a₄ = a+(4-1)d

= a+3d

aur 8th term = a₈ = a+(8-1)d

= a+7d

Ab question ke anusaar-:

⇒ a₄+a₈ = 24

⇒ (a+3d)+(a+7d) =24

⇒ 2a+10d = 24

⇒ a+5d = 12 . . . . . . . . . . . . . . . .(i)

Phir, 6th term = a₆= a+(6-1)d

= a+5d

aur 10th term = a₁₀= a+(10-1)d

= a+9d

Ab question ke anusaar-:

⇒ a₆+a₁₀ = 44

⇒ (a+5d)+(a+9d) = 44

⇒ 2a+14d = 44

⇒ a+7d = 22 . . . . . . . . . . . . . . . .(ii)

Equation (ii) me se Equation (i) ko subtract karne par

a + 7d = 22

a + 5d = 12

- - -

2d = 10

⇒ 2d = 10

⇒ d = 10/2

⇒ d = 5

Equation (i) me d = 5 rakhane par

⇒ a+5×5 = 12

⇒ a+25 = 12

⇒ a = 12-25

⇒ a = -13

Isliye, A.P. ka 1st term = -13, 2nd term = -8 aur 3rd term = -3 Ans.

Ye bhi padhe-:

❒ Prove that medians of a triangle divides each other in the ratio 2:1

❒ Base 100 multiplication tricks for 2 digit numbers

❒ Base method multiplication shortcut trick

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