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If (a_1=12) and it is not (a_n=3a_{n-1}+2) but an AP with (a_n=12+(n-1)d), and (a_9=68), what is (a_{22})?यदि (a_1=12) और (a_{n}=3a_{n-1}+2) नहीं बल्कि AP का (a_n=12+(n-1)d) है तथा (a_9=68) है तो (a_{22}) क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: 159159
Using the AP formula, \(a_9=12+(9-1)d\). Thus, \(68=12+8d\), so \(d=7\). Now \(a_{22}=12+(22-1)\times7=12+147=159\). Choosing 166 would result from using an incorrect common difference instead of 7. Exam tip: first find \(d\) from the given term, then use \(n-1\) for the required term.AP के सूत्र से, \(a_9=12+(9-1)d\)। अतः \(68=12+8d\), इसलिए \(d=7\)। अब \(a_{22}=12+(22-1)\times7=12+147=159\)। विकल्प 166 लेने पर पदांतर 7 के बजाय गलत मान लिया जाएगा। परीक्षा टिप: पहले दिए गए पद से \(d\) ज्ञात करें, फिर आवश्यक पद के लिए \(n-1\) का प्रयोग करें।
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