Question

which term of the AP:3, 15,27,39,... will be 132 more than its 54th term?​

Answer 1

Answer:

65

Step-by-step explanation:

3, 15, 27, 39…….

3, 15, 27, 39…….a

3, 15, 27, 39…….a n

3, 15, 27, 39…….a n

3, 15, 27, 39…….a n =a+(n−1)d

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)12

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1n=65

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1n=65

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1n=65

3, 15, 27, 39…….a n =a+(n−1)da=3,d=15−3=12n=54a 54 =a+(n−1)d=3+(54−1)×12=3+53×12=639Term which is 132 more than its 54th term is – 639 +132 = 771.a n =771771=a+(n−1)d771=3+(n−1)12771=3+(n−1)1264=n−1n=65 Was this ansh helpful

so please mark me brainlest