BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Series

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 1556
Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members
Thanked: 448 times
GMAT Score:650
by theCodeToGMAT » Thu Oct 10, 2013 7:08 am
The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirty-fifth term in the sequence?

A. 2
B. 4
C. 7
D. 9
E. 13

OA [spoiler]{B}[/spoiler]
R A H U L
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 10, 2013 7:22 am
theCodeToGMAT wrote:The first three terms of an infinite sequence are 2, 7, and 22. After the first term, each consecutive term can be obtained by multiplying the previous term by 3 and then adding 1. What is the sum of the tens digit and the units digit of the thirty-fifth term in the sequence?

A. 2
B. 4
C. 7
D. 9
E. 13

OA [spoiler]{B}[/spoiler]
Let's examine a few terms and look for a pattern.
02, 07, 22, 67, 202, 607, 1822, 5467, . . .

So, the pattern goes 02, 07, 22, 67, etc.

This pattern repeats every 4 terms.
So, the 4th term ends in 67
The 8th term ends in 67
The 12th term ends in 67
.
.
.
The 32nd term ends in 67
From here, the 33rd term ends in 02
The 34th term ends in 07
The 35th term ends in [spoiler]22[/spoiler]

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 81
Joined: Tue Jun 11, 2013 10:24 pmFollowed by: 1 member
Thanked: 7 times

by sanjoy18 » Thu Oct 10, 2013 12:15 pm
Lets first term T1, second term T2 ..so on
I found the below pattern
T1= 2 = (5*3^0 -1)/2
T2 = 7 = (5*3^1-1)/2
T3=22= (5*3^2 -1)/2
........................
T35= (5*3^34-1)/2

Now our job is to find the last two digit of 3^34

Last two digit can be found very by binomial expansion
3^34= (3^2)17
=(9)^17
=(10-1)^17
last two digit would be 17c1 *10 -1 = 69

hence last two digit of T35= (69*5-1)/2= 44/2=22

hence answer 2+2=4
hence B
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1