Question

निम्नलिखित समांतर श्रेढि़यों में से प्रत्येक श्रेढ़ी में कितने पद हैं?
(i) 7, 13, 19, . . . , 205
(ii) 18, $15\frac{1}{2}$, 13, . . . , – 47

Answer 1

(i) 7, 13, 19, ....... 205
यहाँ प्रथम पद , a = 7
और सार्व अंतर , d = 13 - 7 = 19 - 13 = 6
माना कि n वां पद = 205

हम जानते हैं कि $a_n=a+(n-1)d$
अब, 205 = 7 + (n - 1) × 6
या, 205 - 7 = 6(n - 1)
या, 198 = 6(n - 1)
या, 198/6 = (n - 1)
या, 33 = (n - 1)
या, n = 34
अतः, दिए समानांतर श्रेणी में 34 पद हैं ।

(ii) 18, $15\frac{1}{2}$, 13, . . . , – 47
यहाँ प्रथम पद , a = 18
और सार्व अंतर , d = 15½ - 18 = -5/2
माना कि n वां पद = -47

हम जानते हैं कि $a_n=a+(n-1)d$
अब, -47 = 18 + (n - 1) × (-5/2)
या, -47 - 18 = (-5/2)(n - 1)
या, -65 = (-5/2)(n - 1)
या, 13 × 2 = (n - 1)
या, 26 = (n - 1)
या, n = 27
अतः, दिए समानांतर श्रेणी में 27 पद हैं ।

Answer 2

Solution :

i ) 7 , 13 , 19 ...., 205 is an A.P

first term = a = 7

common difference= a2 - a1

=> d = 13 - 7

=> d = 6

Let number of term in A.P = n

last term ( l ) = 205

=> a + ( n - 1 )d = 205

=> 7 + ( n - 1 )6 = 205

=> ( n - 1 )6 = 205 - 7

=> ( n - 1 )6 = 198

=> n - 1 = 198/6

=> n - 1 = 33

=> n = 33 + 1 = 34

Therefore ,

Number of terms in given A.P = n = 34
____________________________

ii ) Given A.P : 18, 15½ , 13 , ...., -47

first term = a = 18 ,

common difference = a2 - a1

=> d = 15½ - 18

=> d = -2½ = - 5/2

Let number of terms in A.P = n,

nth term = an = -47 [ given ]

=> 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 = 26

=> n = 26 + 1 = 27

Therefore ,

number of terms in given A.P = n = 27

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