एक समांतर श्रेणी में, तीसरे, चौथे और पाँचवें पदों का योग 12 है। पहले 7 पदों का योग क्या है?

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AAI Junior Assistant (Fire Service) Official Paper (Held On: 15 Nov, 2022 Shift 2)
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  1. 32
  2. 24
  3. 26
  4. 28

Answer (Detailed Solution Below)

Option 4 : 28
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दिया गया है:

समांतर श्रेणी (AP)

तीसरे, चौथे और पाँचवें पदों का योग = 12

प्रयुक्त सूत्र:

AP का nवाँ पद: an = a + (n - 1)d

AP के पहले n पदों का योग: Sn = (n/2)[2a + (n - 1)d]

जहाँ a पहला पद है और d सार्व अंतर है।

गणना:

तीसरा, चौथा और पाँचवाँ पद:

a3 = a + 2d

a4 = a + 3d

a5 = a + 4d

तीसरे, चौथे और पाँचवें पदों का योग = 12:

(a + 2d) + (a + 3d) + (a + 4d) = 12

3a + 9d = 12

a + 3d = 4 ... (1)

पहले 7 पदों का योग (S7):

S7 = (7/2)[2a + (7 - 1)d]

S7 = (7/2)[2a + 6d]

S7 = 7(a + 3d)

समीकरण (1) से a + 3d = 4 रखने पर:

S7 = 7 x 4

S7 = 28

इसलिए, पहले 7 पदों का योग 28 है।

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