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What is the sum of the first (12) terms of the arithmetic progression (45,39,33,\ldots)?समांतर श्रेणी (45,39,33,\ldots) के पहले (12) पदों का योग कितना है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (144)(144)
The twelfth term is (-21), so (S_{12}=\frac{12}{2}(45-21)=144). In decreasing order, keep (d) negative while finding the last term.बारहवाँ पद (-21) है, इसलिए (S_{12}=\frac{12}{2}(45-21)=144)। घटते क्रम में अंतिम पद निकालते समय (d) ऋणात्मक रखें।
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