IBPS RRB Officer Scale-I Prelims practice question
In an arithmetic progression the 5th term is 31 and the 11th term is 67. What is the sum of its first 30 terms?
From a IBPS RRB Officer Scale-I Prelims Quantitative Aptitude practice set · Quantitative Aptitude
Tap an option to check your answer.
Show answer and explanation
Correct answer: (D) 2820
t11 − t5 = (11 − 5)d ⇒ 67 − 31 = 6d ⇒ d = 6. Then a = t5 − (5 − 1)d = 31 − 24 = 7. S30 = 30/2 × [2 × 7 + 29 × 6] = 30/2 × 188 = 2820.
हिंदी में प्रश्न
एक समांतर श्रेढ़ी का 5वाँ पद 31 और 11वाँ पद 67 है। इसके प्रथम 30 पदों का योग क्या है?
उत्तर जाँचने के लिए किसी विकल्प पर टैप करें।
उत्तर और व्याख्या देखें
सही उत्तर: (D) 2820
t11 − t5 = (11 − 5)d ⇒ 67 − 31 = 6d ⇒ d = 6। तब a = t5 − (5 − 1)d = 31 − 24 = 7। S30 = 30/2 × [2 × 7 + 29 × 6] = 30/2 × 188 = 2820।
More solved Quantitative Aptitude questions
- 7 years ago, the average age of a group of six colleagues of 6 members was 30 years. No member has left or joined the group of six…
- Find the next term: 3, 10, 29, 66, ?
- Deepak holds a promissory note of Rs. 3120 due 4 years from now. If money is worth 6% per annum, what is the true discount on this…
- At the same constant speed, 3 patrol jeeps running 9 hours a day cover 1080 km in 4 days. Find the number of days needed for 6 such patrol…
- The diagonal of a cuboid is 11 cm. If its length is 6 cm and its height is 7 cm, find its breadth.
- A jeweller calculates a design difference as 5√2 − 2√5 (in grams). Using √2 = 1.414 and √5 = 2.236, find this value to 3 decimal places.